Advanced Quantum Field Theory

Home , Gauge principle, Scalar field theory

Roberto Casalbuoni Dipartimento di Fisica Universit`a di Firenze

Lezioni date all’Universita’ di Firenze nell’a.a. 2004/2005. Contents

Index...... 1

1 Notations and Conventions 4 1.1Units...... 4 1.2RelativityandTensors...... 6 1.3TheNoether’stheoremforrelativisticﬁelds...... 6 1.4FieldQuantization...... 9 1.4.1 TheRealScalarField...... 10 1.4.2 TheChargedScalarField...... 12 1.4.3 TheDiracField...... 14 1.4.4 TheElectromagneticField...... 17 1.5PerturbationTheory...... 27 1.5.1 TheScatteringMatrix...... 27 1.5.2 Wick’stheorem...... 29 1.5.3 Feynmandiagramsinmomentumspace...... 30 1.5.4 Thecross-section...... 33

2 One-loop renormalization 36 2.1DivergencesoftheFeynmanintegrals...... 36 2.2Higherordercorrections...... 38 2.3Theanalysisbycounterterms...... 45 2.4DimensionalregularizationoftheFeynmanintegrals...... 52 2.5Integrationinarbitrarydimensions...... 54 2.6OneloopregularizationofQED...... 56 2.7Onelooprenormalization...... 63

3 Vacuum expectation values and the S matrix 73 3.1In-andOut-states...... 73 3.2 The S matrix...... 79 3.3Thereductionformalism...... 81

4 Path integral formulation of quantum mechanics 85 4.1Feynman’sformulationofquantummechanics...... 85 4.2Pathintegralinconﬁgurationspace...... 88

1 4.3Thephysicalinterpretationofthepathintegral...... 91 4.4Thefreeparticle...... 97 4.5Thecaseofaquadraticaction...... 99 4.6Functionalformalism...... 104 4.7Generalpropertiesofthepathintegral...... 107 4.8ThegeneratingfunctionaloftheGreen’sfunctions...... 113 4.9 The Green’s functions for the harmonic oscillator ...... 116

5 The path integral in ﬁeld theory 123 5.1Thepathintegralforafreescalarﬁeld...... 123 5.2 The generating functional of the connected Green’s functions . . . . . 127 5.3 The perturbative expansion for the theory λϕ4 ...... 129 5.4TheFeynman’srulesinmomentumspace...... 135 5.5 Power counting in λϕ4 ...... 137 5.6 Regularization in λϕ4 ...... 142 5.7 Renormalization in the theory λϕ4 ...... 145

6 The renormalization group 152 6.1Therenormalizationgroupequations...... 152 6.2Renormalizationconditions...... 156 6.3ApplicationtoQED...... 160 6.4Propertiesoftherenormalizationgroupequations...... 161 6.5 The coeﬃcients of the renormalization group equations and the renormalizationconditions...... 167

7 The path integral for Fermi ﬁelds 169 7.1 Fermionic oscillators and Grassmann algebras ...... 169 7.2IntegrationoverGrassmannvariables...... 174 7.3Thepathintegralforthefermionictheories...... 178

8 The quantization of the gauge ﬁelds 180 8.1QEDasagaugetheory...... 180 8.2Non-abeliangaugetheories...... 182 8.3Pathintegralquantizationofthegaugetheories...... 186 8.4PathintegralquantizationofQED...... 196 8.5 Path integral quantization of the non-abelian gauge theories . . . . . 200 8.6 The β-functioninnon-abeliangaugetheories...... 204

9 Spontaneous symmetry breaking 213 9.1 The linear σ-model...... 213 9.2Spontaneoussymmetrybreaking...... 219 9.3TheGoldstonetheorem...... 223 9.4TheHiggsmechanism...... 225 9.5 Quantization of a spontaneously broken gauge theory in the Rξ-gauge 230

2 9.6 ξ-cancellationinperturbationtheory...... 235

10 The Standard model of the electroweak interactions 239 10.1TheStandardModeloftheelectroweakinteractions...... 239 10.2TheHiggssectorintheStandardModel...... 244 10.3 The electroweak interactions of quarks and the Kobayashi-Maskawa- Cabibbomatrix...... 249 10.4TheparametersoftheSM...... 258 10.5Wardidentitiesandanomalies...... 259

A 265 A.1Propertiesoftherealantisymmetricmatrices...... 265

3 Chapter 1

Notations and Conventions

1.1 Units

In quantum relativistic theories the two fundamental constants c e /h, the light velocity and the Planck constant respectively, appear everywhere. Therefore it is convenient to choose a unit system where their numerical value is given by

c = /h =1 (1.1)

For the electromagnetism we will use the Heaviside-Lorentz system, where we take also 0 =1 (1.2) 2 From the relation 0µ0 =1/c it follows

µ0 =1 (1.3)

In these units the Coulomb force is given by e e 1 |F | 1 2 = 2 (1.4) 4π |x1 − x2| and the Maxwell equations appear without any visible constant. For instance the gauss law is ∇· E = ρ (1.5) dove ρ is the charge density. The dimensionless ﬁne structure constant

e2 α = / (1.6) 4π0hc is given by e2 α = (1.7) 4π

4 Any physical quantity can be expressed equivalently by using as fundamental unit energy, mass, lenght or time in an equivalent fashion. In fact from our choice the following equivalence relations follow

ct ≈ =⇒ time ≈ lenght E ≈ mc2 =⇒ energy ≈ mass E ≈ pv =⇒ energy ≈ momentum Et ≈ /h =⇒ energy ≈ (time)−1 ≈ (lenght)−1 (1.8)

In practice, it is enough to notice that the product ch/ has dimensions [E· ]. Therefore

ch/ =3· 108 mt · sec−1 · 1.05 · 10−34 J · sec = 3.15 · 10−26 J · mt (1.9)

Recalling that 1eV=e · 1=1.602 · 10−19 J (1.10) it follows 3.15 · 10−26 ch/ = MeV · mt = 197 MeV · fermi (1.11) 1.6 · 10−13 From which 1MeV−1 = 197 fermi (1.12) Using this relation we can easily convert a quantity given in MeV (the typical unit used in elementary particle physics) in fermi. For instance, using the fact that also the elementary particle masses are usually given in MeV , the wave lenght of an electron is given by · e 1 ≈ 1 ≈ 200 MeV fermi ≈ λCompton = 400 fermi (1.13) me 0.5 MeV 0.5 MeV Therefore the approximate relation to keep in mind is 1 = 200 MeV · fermi. Fur- thermore, using c =3· 1023 fermi · sec−1 (1.14) we get 1fermi=3.3 · 10−24 sec (1.15) and 1MeV−1 =6.58 · 10−22 sec (1.16) Also, using 1 barn = 10−24 cm2 (1.17) it follows from (1.12) 1GeV−2 =0.389 mbarn (1.18)

5 1.2 Relativity and Tensors

Our conventions are as follows: the metric tensor gµν is diagonal with eigenvalues (+1, −1, −1, −1). The position and momentum four-vectors are given by

xµ =(t, x),pµ =(E,p),µ=0, 1, 2, 3 (1.19) where x and p are the three-dimensional position and momentum. The scalar product between two four-vectors is given by

µ ν µ µ 0 0 a · b = a b gµν = aµb = a bµ = a b − a ·b (1.20) where the indices have been lowered by

ν aµ = gµν a (1.21)

µν and can be raised by using the inverse metric tensor g ≡ gµν . The four-gradient is deﬁned as ∂ ∂ ∂ ∂µ = = , = ∂t, ∇ (1.22) ∂xµ ∂t ∂x The four-momentum operator in position space is

µ ∂ p → i =(i∂t, −i∇ ) (1.23) ∂xµ

The following relations will be useful

2 µ →− ∂ ∂ − p = pµp µ = (1.24) ∂x ∂xµ

x · p = Et − p · x (1.25)

1.3 The Noether’s theorem for relativistic ﬁelds

We will now review the Noether’s theorem. This allows to relate symmetries of the action with conserved quantities. More precisely, given a transformation involving both the ﬁelds and the coordinates, if it happens that the action is invariant under this transformation, then a conservation law follows. When the transformations are limited to the ﬁelds one speaks about internal transformations.Whenboth types of transformations are involved the total variation of a local quantity F (x) (that is a function of the space-time point) is given by

∆F (x)=F˜(x) − F (x)=F˜(x + δx) − F (x) ∼ µ ∂F(x) = F˜(x) − F (x)+δx (1.26) ∂xµ

6 The total variation ∆ keeps into account both the variation of the reference frame and the form variation of F . It is then convenient to deﬁne a local variation δF, depending only on the form variation δF(x)=F˜(x) − F (x) (1.27) Then we get ∂F(x) ∆F (x)=δF(x)+δxµ (1.28) ∂xµ Let us now start form a generic four-dimensional action S = d4x L(φi,x),i=1,...,N (1.29) V µ and let us consider a generic variation of the ﬁelds and of the coordinates, x = xµ + δxµ ∂φ ∆φi(x)=φ˜i(x) − φi(x) ≈ δφi(x)+δxµ (1.30) ∂xµ If the action is invariant under the transformation, then

S˜V = SV (1.31) This gives rise to the conservation equation L L L µ ∂ i − ν ∂ i ∂µ δx + i ∆φ δx i φ,ν = 0 (1.32) ∂φ,µ ∂φ,µ This is the general result expressing the local conservation of the quantity in parenthesis. According to the choice one does for the variations δxµ and ∆φi,andof the corresponding symmetries of the action, one gets diﬀerent kind of conserved quantities. Let us start with an action invariant under space and time translations. In the case we take δxµ = aµ with aµ independent on x e∆φi = 0. From the general result in eq. (1.32) we get the following local conservation law L µ ∂ i −L µ µ Tν = i φ,ν gν ,∂µTν = 0 (1.33) ∂φ,µ

Tµν is called the energy-momentum tensor of the system. From its local conservation we get four constant of motion 3 0 Pµ = d xTµ (1.34)

Pµ is the four-momentum of the system. In the case of internal symmetries we take δxµ = 0. The conserved current will be

µ ∂L i ∂L i µ J = i ∆φ = i δφ ,∂µJ = 0 (1.35) ∂φ,µ ∂φ,µ

7 with an associated constant of motion given by Q = d3xJ 0 (1.36)

In general, if the system has more that one internal symmetry, we may have more that one conserved charge Q, that is we have a conserved charge for any ∆. The last case we will consider is the invariance with respect to Lorentz transformations. Let us recall that they are deﬁned as the transformations leaving invariant the norm of a four-vector x2 = x2 (1.37) For an inﬁnitesimal transformation

ν ν xµ =Λµν x ≈ xµ + µν x (1.38) with µν = −νµ (1.39) We see that the number of independent parameters characterizing a Lorentz transformation is six. As well known, three of them correspond to spatial rotations, whereas the remaining three correspond to Lorentz boosts. In general, the relativistic fields are chosen to belong to a representation of the Lorentz group ( for instance the Klein-Gordon field belongs to the scalar representation). This means that under a Lorentz transformation the components of the field mix together, as, for instance, a vector field does under rotations. Therefore, the transformation law of the fields φi under an infinitesimal Lorentz transformation can be written as 1 ∆φi = − Σij µν φj (1.40) 2 µν where we have required that the transformation of the fields is of first order in the Lorentz parameters µν . The coefficients Σµν (antisymmetric in the indices (µ, ν)) define a matrix in the indices (i, j) which can be shown to be the representative of the infinitesimal generators of the Lorentz group in the field representation. Using this equation and the expression for δxµ we get the local conservation law ∂L i µ νρ 1 ∂L ij νρ j 0=∂µ φ −Lg xρ + Σ φ ∂φi ,ν ν 2 ∂φi νρ ,µ ,µ L 1 νρ µ − µ ∂ ij j = ∂µ Tν xρ Tρ xν + i Σνρφ (1.41) 2 ∂φ,µ and defining (watch at the change of sign) L Mµ µ − µ − ∂ ij j ρν = xρTν xν Tρ i Σρνφ (1.42) ∂φ,µ

8 it follows the existence of six locally conserved currents (one for each Lorentz transformation) µ ∂µMνρ = 0 (1.43) and consequently six constants of motion (notice that the lower indices are antisymmetric) 3 0 Mνρ = d xMνρ (1.44)

Three of these constants (the ones with ν and ρ assuming spatial values) are nothing but the components of the angular momentum of the ﬁeld.

1.4 Field Quantization

The quantization procedure for a ﬁeld goes through the following steps:

• construction of the lagrangian density and determination of the canonical momentum density ∂L Πα(x)= (1.45) ∂φα(x) where the index α describes both the spin and internal degrees of freedom

• quantization through the requirement of the equal time canonical commutation relations for a spin integer ﬁeld or through canonical anticommutation relations for half-integer ﬁelds

3 [φα(x, t), Πβ(y, t)]± = iδαβ δ (x − y) (1.46)

[φα(x, t),φβ(y,t)]± =0, [Πα(x, t), Πβ(y, t)]± = 0 (1.47)

In the case of a free ﬁeld, or for an interacting ﬁeld in the interaction picture (and therefore evolving with the free hamiltonian), we can add the following steps

• expansion of φα(x) in terms of a complete set of solutions of the Klein-Gordon equation, allowing the deﬁnition of creation and annihilation operators;

• construction of the Fock space through the creation and annihilation operators.

In the following we will give the main results about the quantization of the scalar, the Dirac and the electromagnetic ﬁeld.

9 1.4.1 The Real Scalar Field The relativistic real scalar free ﬁeld φ(x) obeys the Klein-Gordon equation

( + m2)φ(x) = 0 (1.48) which can be derived from the variation of the action S = d4xL (1.49) where L is the lagrangian density 1 µ 2 2 L = ∂µφ∂ φ − m φ (1.50) 2 giving rise to the canonical momentum ∂L Π= = φ˙ (1.51) ∂φ˙ From this we get the equal time canonical commutation relations

[φ(x, t), φ˙(y, t)] = iδ3(x − y), [φ(x, t),φ(y, t)] = [φ˙(x, t), φ˙(y, t)] = 0 (1.52)

By using the box normalization, a complete set of solutions of the Klein-Gordon equation is given by

√1 √1 −ikx ∗ √1 √1 ikx fk(x)= e ,fk (x)= e (1.53) V 2ωk V 2ωk corresponding respectively to positive and negative energies, with 2 2 k0 = ωk = |k| + m (1.54) and 2π k = n (1.55) L with n = n1i1 + n2i2 + n3i3,ni ∈ ZZ (1.56) and L being the side of the quantization box (V = L3). For any two solutions f and g of the Klein-Gordon equation, the following quantity is a constant of motion and deﬁnes a scalar product (not positive deﬁnite) 3 ∗ (−) f|g = i d xf ∂t g (1.57) where ∗ (−) ∗ ∗ f ∂t g = f g˙ − f˙ g (1.58)

10 The solutions of eq. (1.53) are orthogonal with respect to this scalar product, for instance 3

| ≡ fk fk = δk,k δni,ni (1.59) i=1 We can go to the continuum normalization by the substitution 1 1 √ → (1.60) V (2π)3 and sending the Kr¨onecker delta function into the Dirac delta function. The expansion of the ﬁeld in terms of these solutions is then (in the continuum) 3 ∗ † φ(x)= d k[fk(x)a(k)+fk (x)a (k)] (1.61) or, more explicitly 1 − † φ(x)= d3k [a(k)e ikx + a (k)eikx] (1.62) (2π)32ωk using the orthogonality properties of the solutions one can invert this expression 3 ∗ (−) † 3 (−) a(k)=i d xfk (x)∂t φ(x),a(k)=i d xφ(x)∂t fk(x) (1.63) and obtain the commutation relations

[a(k),a†(k)] = δ3(k − k) (1.64)

[a(k),a(k)] = [a†(k),a†(k)] = 0 (1.65) From these equations one constructs the Fock space starting from the vacuum state whichisdeﬁnedby a(k)|0 = 0 (1.66) for any k.Thestates † † a (k1) ···a (kn)|0 (1.67) 2 2 describe n Bose particles of equal mass m = ki , and of total four-momentum k = k1 + ···kn. This follows by recalling the expression (1.33) for the energy- momentum tensor, that for the Klein-Gordon ﬁeld is 1 ρ 2 2 Tµν = ∂µφ∂ν φ − ∂ρφ∂ φ − m φ gµν (1.68) 2 from which 1 P 0 = H = d3xT 00 = d3x Π2 + |∇ φ|2 + m2φ2 (1.69) 2

11 ∂φ P i = d3xT 0i = − d3xΠ , (P = − d3xΠ∇ φ) (1.70) ∂xi and for the normal ordered operator Pµ we get (the normal ordering is deﬁned by putting all the creation operators to the left of the annihilation operators) 3 † : Pµ := d kkµa (k)a(k) (1.71) implying † † : Pµ : a (k)|0 = kµa (k)|0 (1.72) The propagator for the scalar ﬁeld is given by

0|T (φ(x)φ(y))|0 = −i∆F (x − y) (1.73) where 4 d k −ikx ∆F (x)=− e ∆F (k) (1.74) (2π)4 and 1 ∆F (k)= (1.75) k2 − m2 + i

1.4.2 The Charged Scalar Field The charged scalar ﬁeld can be described in terms of two real scalar ﬁelds of equal mass 2 1 µ 2 2 L = (∂µφi)(∂ φi) − m φ (1.76) 2 i i=1 and we can write immediately the canonical commutation relations

3 [φi(x, t), φ˙j(y, t)] = iδijδ (x − y) (1.77)

[φi(x, t),φj(y, t)] = [φ˙i(x, t), φ˙j(y, t)] = 0 (1.78) The charged ﬁeld is better understood in a complex basis

1 † 1 φ = √ (φ1 + iφ2),φ= √ (φ1 − iφ2) (1.79) 2 2 The lagrangian density becomes

† µ 2 † L = ∂µφ ∂ φ − m φ φ (1.80)

The theory is invariant under the phase transformation φ → eiαφ, and therefore it admits the conserved current (see eq. (1.35)) † † jµ = i (∂µφ)φ − (∂µφ )φ (1.81)

12 with the corresponding constant of motion 3 † (−) Q = i d xφ∂t φ (1.82)

The commutation relations in the new basis are given by

[φ(x, t), φ˙†(y, t)] = iδ3(x − y) (1.83) and

[φ(x, t),φ(y, t)] = [φ†(x, t),φ†(y, t)] = 0 [φ˙(x, t), φ˙(y, t)] = [φ˙†(x, t), φ˙†(y, t)] = 0 (1.84)

Let us notice that these commutation relations could have also been obtained directly from the lagrangian (1.80), since

∂L † ∂L Πφ = = φ˙ , Πφ† = = φ˙ (1.85) ∂φ˙ ∂φ˙† Using the expansion for the real ﬁelds 3 ∗ † φi(x)= d k fk(x)ai(k)+fk (x)a i(k) (1.86) we get 3 √1 ∗ √1 † † φ(x)= d k fk(x) (a1(k)+ia2(k)) + f (x) (a 1(k)+ia 2(k)) (1.87) 2 k 2 Introducing the combinations

1 1 a(k)=√ (a1(k)+ia2(k)),b(k)=√ (a1(k) − ia2(k)) (1.88) 2 2 it follows 3 ∗ † φ(x)= d k fk(x)a(k)+fk (x)b (k) † 3 ∗ † φ (x)= d k fk(x)b(k)+fk (x)a (k) (1.89) from which we can evaluate the commutation relations for the creation and annihilation operators in the complex basis

[a(k),a†(k)] = [b(k),b†(k)] = δ3(k − k) (1.90)

[a(k),b(k)] = [a(k),b†(k)] = 0 (1.91)

13 We get also

2 3 † 3 † † : Pµ := d kkµ a i(k)ai(k)= d kkµ a (k)a(k)+b (k)b(k) (1.92) i=1

Therefore the operators a†(k)eb†(k) both create particles states with momentum † k, as the original operators a i. For the normal ordered charge operator we get : Q := d3k a†(k)a(k) − b†(k)b(k) (1.93) showing explicitly that a† and b† create particles of charge +1 and −1 respectively. The only non vanishing propagator for the scalar ﬁeld is given by

† 0|T (φ(x)φ (y))|0 = −i∆F (x − y) (1.94)

1.4.3 The Dirac Field The Dirac ﬁeld is described by the action S = d4x ψ¯(i∂ˆ − m)ψ (1.95) V where we have used the following notations for four-vectors contracted with the γ matrices µ vˆ ≡ vµγ (1.96) The canonical momenta result to be

∂L † ∂L Πψ = = iψ , Πψ† = = 0 (1.97) ∂ψ˙ ∂ψ˙† The canonical momenta do not depend on the velocities. In principle, this creates a problem in going to the hamiltonian formalism. In fact a rigorous treatment requires an extension of the classical hamiltonian approach which was performed by Dirac himself. In this particular case, the result one gets is the same as proceeding in a naive way. For this reason we will avoid to describe this extension, and we will proceed as in the standard case. From the general expression for the energy momentum tensor (see eq. 1.33) we get

µ µ µ Tν = iψγ¯ ψ,ν − gν (ψ¯(i∂ˆ − m)ψ) (1.98) and using the Dirac equation µ µ Tν = iψγ¯ ψ,ν (1.99) The momentum of the ﬁeld is given by P k = d3xT0k =⇒ P = −i d3xψ†∇ ψ (1.100)

14 and the hamiltonian 3 00 3 † H = d xT =⇒ H = i d xψ∂tψ (1.101)

For a Lorentz transformation, the ﬁeld variation is deﬁned by 1 ∆φi = − Σij µν φj (1.102) 2 µν In the Dirac case one has

i µν ∆ψ(x)=ψ (x ) − ψ(x)=− σµν ψ(x) (1.103) 4 from which i 1 Σµν = σµν = − [γµ,γν] (1.104) 2 4 Therefore the angular momentum density (see eq. (1.42)) is µ µ i µ 1 M = iψγ¯ xρ∂ν − xν∂ρ − σρν ψ = iψγ¯ xρ∂ν − xν ∂ρ + [γρ,γν] ψ ρν 2 4 (1.105) By taking the spatial components we obtain 1 J =(M 23,M31,M12)= d3xψ† −ix ∧ ∇ + σ ⊗ 1 ψ (1.106) 2 2 where 1 is the identity matrix in 2 dimensions, and we have deﬁned 2 σ 0 σ ⊗ 1 = (1.107) 2 0 σ

The expression of J shows the decomposition of the total angular momentum in the orbital and in the spin part. The theory has a further conserved quantity, the current ψγ¯ µψ resulting from the phase invariance ψ → eiαψ. The decomposition of the Dirac ﬁeld in plane waves is obtained by using the spinors u(p, ±n)ev(p, ±n) solutions of the Dirac equation in momentum space for positive and negative energies respectively. We write 3 d p m − † ψ(x)= b(p, n)u(p, n)e ipx + d (p, n)v(p, n)eipx (1.108) 3 ±n (2π) Ep 3 † d p m −ipx † ipx ψ (x)= d(p, n)¯v(p, n)e + b (p, n)¯u(p, n)e γ0 3 ±n (2π) Ep (1.109) where Ep = |p|2 + m2. We will collect here the various properties of the spinors. The four-vector nµ is the one that identiﬁes the direction of the spin quantization in the rest frame. For instance, by quantizing the spin along the third axis one takes µ µ nµ in the rest frame as nR =(0, 0, 0, 1). Then n is obtained by boosting from the rest frame to the frame where the particle has four-momentum pµ.

15 • Dirac equation

(ˆp − m)u(p, n)=¯u(p, n)(ˆp − m)=0 (ˆp + m)v(p, n)=¯v(p, n)(ˆp + m) = 0 (1.110)

• Orthogonality

u¯(p, n)u(p, n )=−v¯(p, n)v(p, n )=δnn † † Ep u (p, n)u(p, n )=v (p, n)v(p, n )= δnn m v¯(p, n)u(p, n)=v†(p, n)u(˜p, n) = 0 (1.111)

µ µ where, if p =(Ep,p), thenp ˜ =(Ep, −p).

• Completeness pˆ + m u(p, n)¯u(p, n)= ±n 2m pˆ − m v(p, n)¯v(p, n)= (1.112) ±n 2m

The Dirac ﬁeld is quantized by canonical anticommutation relations in order to satisfy the Pauli principle † † [ψ(x, t),ψ(y, t)]+ = ψ (x, t),ψ (y, t) + = 0 (1.113) and 3 − [Πψ(x, t),ψ(y, t)]+ = iδ (x y) (1.114) or † 3 − ψ(x, t),ψ (y, t) + = δ (x y) (1.115) The normal ordered four-momentum is given by 3 † † : Pµ := d ppµ[b (p, n)b(p, n)+d (p, n)d(p, n)] (1.116) ±n

If we couple the Dirac field to the electromagnetism through the minimal substitution we find that the free action (1.95) becomes S = d4xψ¯(i∂ˆ − eAˆ − m)ψ (1.117) V Therefore the electromagnetic field is coupled to the conserved current

jµ = eψγ¯ µψ (1.118)

16 This forces us to say that the integral of the fourth component of the current should be the charge operator. In fact, we ﬁnd : Q :=e d3xψ†ψ = d3pe b†(p, n)b(p, n) − d†(p, n)d(p, n) (1.119) ±n

The expressions for the momentum angular momentum and charge operators show that the operators b†(p)andd†(p) create out of the vacuum particles of spin 1/2, four-momentum pµ and charge e and −e respectively. The propagator for the Dirac ﬁeld is given by

0|T (ψα(x)ψ¯β(y))|0 = iSF (x − y)αβ (1.120) where 4 d k −ikx SF (x)=−(i∂ˆ + m)∆F (x)= e SF (k) (1.121) (2π)4 and kˆ + m 1 SF (k)= = (1.122) k2 − m2 + i kˆ − m + i

1.4.4 The Electromagnetic Field The lagrangian density for the free electromagnetic ﬁeld, expressed in terms of the four-vector potential is 1 µν L = − Fµν F (1.123) 4 where Fµν = ∂µAν − ∂νAµ (1.124) and ∂A E = −∇ A0 − , B = ∇∧ A (1.125) ∂t from which E =(F 10,F20,F30), B =(−F 23, −F 31, −F 12) (1.126) The resulting equations of motion are

ν Aµ − ∂µ(∂ Aν ) = 0 (1.127)

The potentials are deﬁned up to a gauge transformation

Aµ(x) → Aµ(x)=Aµ(x)+∂µΛ(x) (1.128)

In fact, Aµ and Aµ satisfy the same equations of motion and give rise to the same electromagnetic ﬁeld. It is possible to use the gauge invariance to require some

17 particular condition on the field Aµ. For instance, we can perform a gauge transformation in such a way that the transformed field satisfies

µ ∂µA = 0 (1.129)

This is called the Lorentz gauge. Or we can choose a gauge such that A0 = ∇·A = 0 (1.130)

This is called the Coulomb gauge. Since in this gauge all the gauge freedom is completely ﬁxed (in contrast with the Lorentz gauge), it is easy to count the independent degrees of freedom of the electromagnetic ﬁeld. From the equations (1.130) we see that Aµ has only two degrees of freedom. Another way of showing that there are only two independent degrees of freedom is through the equations of motion. Let us consider the four dimensional Fourier transform of Aµ(x) 4 ikx Aµ(x)= d ke Aµ(k) (1.131)

Substituting this expression in the equations of motion we get

2 ν −k Aµ(k)+kµ(k Aν (k)) = 0 (1.132)

Let us now decompose Aµ(k) in terms of four independent four vectors, which can µ µ λ be chosen as k =(E,k), k˜ =(E,−k), and two further four vectors eµ(k), λ =1, 2, orthogonal to kµ µ λ k eµ =0,λ=1, 2 (1.133)

The decomposition of Aµ(k)reads

λ Aµ(k)=aλ(k)eµ + b(k)kµ + c(k)k˜µ (1.134)

From the equations of motion we get

2 λ 2 −k (aλeµ + bkµ + ck˜µ)+kµ(bk + c(k · k˜)) = 0 (1.135)

The term in b(k) cancels, therefore it is left undetermined by the equations of motion. For the other quantities we have

2 k aλ(k)=c(k) = 0 (1.136)

The arbitrariness of b(k) is a consequence of the gauge invariance. In fact if we gauge transform Aµ(x) Aµ(x) → Aµ(x)+∂µΛ(x) (1.137) then Aµ(k) → Aµ(k)+ikµΛ(k) (1.138)

18 where Λ(x)= d4keikxΛ(k) (1.139)

Since the gauge transformation amounts to a translation in b(k) by an arbitrary function of k, we can always to choose it equal to zero. Therefore we are left with the two degrees of freedom described by the amplitudes aλ(k), λ =1, 2. Furthermore these amplitudes are different from zero only if the dispersion relation k2 =0is satisfied. This shows that the corresponding quanta will have zero mass. With the choice b(k)=0,thefieldAµ(k) becomes

λ Aµ(k)=aλ(k)eµ(k) (1.140)

µ showing that k Aµ(k) = 0. Therefore the choice b(k)=0isequivalenttothechoice of the Lorentz gauge. Let us consider now the quantization of this theory. Due to the lack of manifest covariance of the Coulomb gauge we will discuss here the quantization in the Lorentz gauge, where however we will encounter other problems. If we want to maintain the explicit covariance of the theory we have to require non trivial commutation relations for all the component of the ﬁeld. That is

ν ν 3 [Aµ(x, t), Π (y, t)] = igµδ (x − y) (1.141)

µ ν [Aµ(x, t),Aν(y, t)] = [Π (x, t), Π (y, t)] = 0 (1.142) with ∂L Πµ = (1.143) ∂A˙µ To evaluate the conjugated momenta is better to write the lagrangian density (see eq. (1.123) in the following form

1 µ,ν ν,µ 1 µ,ν 1 ν,µ L = − [Aµ,ν − Aν,µ][A − A ]=− Aµ,ν A + Aµ,ν A (1.144) 4 2 2 Therefore ∂L = −Aµ,ν + Aν,µ = F µν (1.145) ∂Aµ,ν implying ∂L Πµ = = F µ0 (1.146) ∂A˙µ It follows ∂L Π0 = = 0 (1.147) ˙ ∂A0 We see that it is impossible to satisfy the condition

0 3 [A0(x, t), Π (y, t)] = iδ (x − y) (1.148)

19 We can try to ﬁnd a solution to this problem modifying the lagrangian density in such a way that Π0 = 0. But doing so we will not recover the Maxwell equation. However we can take advantage of the gauge symmetry, modifying the lagrangian density in such a way to recover the equations of motion in a particular gauge. For instance, in the Lorentz gauge we have

Aµ(x) = 0 (1.149) and this equation can be obtained by the lagrangian density

1 µ,ν L = − Aµ,ν A (1.150) 2 (just think to the Klein-Gordon case). The minus sign is necessary to recover a positive hamiltonian density. We now express this lagrangian density in terms of the gauge invariant one, given in eq. (1.123). To this end we observe that the diﬀerence between the two lagrangian densities is nothing but the second term of eq. (1.144) 1 ν,µ µ 1 ν 1 µ ν Aµ,ν A = ∂ Aµ,ν A − (∂ Aµ,ν )A 2 2 2 µ 1 ν ν 1 µ 1 µ 2 = ∂ Aµ,ν A − ∂ (∂ Aµ)Aν + (∂ Aµ) (1.151) 2 2 2 Then, up to a four divergence, we can write the new lagrangian density in the form

1 µν 1 µ 2 L = − Fµν F − (∂ Aµ) (1.152) 4 2 One can check that this form gives the correct equations of motion. In fact from

∂L µ,ν ν,µ µν λ ∂L = −A + A − g (∂ Aλ), = 0 (1.153) ∂Aµ,ν ∂Aµ we get µ µ ν µ λ µ 0=−A + ∂ (∂ Aν) − ∂ (∂ Aλ)=−A (1.154) the term 1 µ 2 − (∂ Aµ) (1.155) 2 which is not gauge invariant, is called the gauge ﬁxing term. More generally, we could add to the original lagrangian density a term of the form

λ µ 2 − (∂ Aµ) (1.156) 2 The corresponding equations of motion would be

µ λ Aµ − (1 − λ)∂ (∂ Aλ) = 0 (1.157)

20 In the following we will use λ = 1. From eq. (1.153) we see that L 0 ∂ µ Π = = −∂ Aµ (1.158) ∂A˙0 In the Lorentz gauge we ﬁnd again Π0 = 0. To avoid the corresponding problem µ we can ask that ∂ Aµ = 0 does not hold as an operator condition, but rather as a condition upon the physical states

µ phys|∂ Aµ|phys = 0 (1.159)

The price to pay to quantize the theory in a covariant way is to work in a Hilbert space much bigger than the physical one. The physical states span a subspace which is deﬁned by the previous relation. A further bonus is that in this way one has to do with local commutation relations. On the contrary, in the Coulomb gauge, one needs to introduce non local commutation relations for the canonical variables. We will come back later to the condition (1.159). Since we don’t have to worry any more about the operator condition Π0 =0,we can proceed with our program of canonical quantization. The canonical momentum densities are L µ ∂ µ0 µ0 λ Π = = F − g (∂ Aλ) (1.160) ∂A˙ µ or, explicitly

0 λ Π = −∂ Aλ = −A˙ 0 − ∇·A i i 0 i i i 0 Π = ∂ A0 − ∂ A = −A˙ + ∂ A (1.161) Since the spatial gradient of the ﬁeld commutes with the ﬁeld itself at equal time, the canonical commutator (1.141) gives rise to

3 [Aµ(x, t), A˙ ν(y,t)] = −igµν δ (x − y) (1.162)

To get the quanta of the ﬁeld we look for plane wave solutions of the wave equation. We need four independent four vectors in order to expand the solutions in the momentum space. In a given frame, let us consider the unit four vector which deﬁnes the time axis. This must be a time-like vector, n2 = 1, and we will choose 0 µ (λ) n > 0. For instance, n =(1, 0, 0, 0). Then we take two four vectors µ , λ =1, 2,in the plane orthogonal to nµ and kµ.Noticethatnowk2 = 0,since we are considering solutions of the wave equation. Therefore

µ (λ) µ (λ) k µ = n µ =0,λ=1, 2 (1.163)

(λ) µ The four vectors µ , being orthogonal to n are space-like, then they will be chosen orthogonal and normalized in the following way µ (λ) (λ) µ = −δλλ (1.164)

21 Next, we deﬁne a unit space-like four vector, orthogonal to nµ and lying in the plane (k, n) . nµ(3) µ = 0 (1.165) with µ (3) (3) µ = −1 (1.166) (3) (λ) By construction µ is orthogonal to µ . This four vector is completely ﬁxed by the previous conditions, and we get

kµ − (n · k)nµ (3) = (1.167) µ (n · k) As a last unit four vector we choose nµ

(0) µ = nµ (1.168) These four vectors are orthonormal µ (λ) (λ) λλ µ = g (1.169) and linearly independent. Then they satisfy the completeness relation

(λ) (λ) µ ν gλλ = gµν (1.170)

In the frame where nµ =(1,0) and kµ =(k, 0, 0,k), we have µ µ µ (1) =(0, 1, 0, 0),(2) =(0, 0, 1, 0),(3) =(0, 0, 0, 1) (1.171)

The plane wave expansion of Aµ is 3 3 d k (λ) −ikx † ikx Aµ(x)= µ (k) aλ(k)e + aλ(k)e (1.172) k 3 2ω (2π) λ=0 where we have included the hermiticity condition for Aµ(x). For any fixed µ,this expansion is the same as the one that we wrote for the Klein-Gordon field, with the (λ) substitution µ aλ(k) → a(k). Then, from eq. (1.63) (λ) 3 ∗ (−) µ (k)aλ(k)=i d xfk (x)∂t Aµ(x) (1.173) with the functions fk(x) defined as in eq. (1.53). Using the orthogonality of the µ (λ ) ’s we find λ µ ∗ (−) 3 ( ) aλ(k)=igλλ d x (k)fk (x)∂t Aµ(x) (1.174) and analogously µ † 3 (λ) (−) aλ(k)=igλλ d x (k)Aµ(x)∂t fk(x) (1.175)

22 From these expressions we can evaluate the commutator † [aλ(k),a (k )] = λ µ ν 3 3 ∗ (λ) (λ) 3 − ˙ − = d xdy fk (x)fk (y) igµν gλλ (k) (k )gλ λ δ (x y) µ ν ˙∗ (λ) (λ) 3 − f (x)fk (y) −igµν gλλ (k) (k )gλλ δ (x − y) k 3 ∗ (−) − = d xfk (x)i∂t fk (x)gλλ (1.176) and using the orthogonality relations for the functions f)k † 3 [aλ(k),aλ (k )] = −gλλ δ (k − k ) (1.177) Analogously † † [aλ(k),aλ (k )] = [aλ(k),aλ (k )] = 0 (1.178)

Therefore the states with λ = 0 have negative norm. This problem does not come out completely unexpected. In fact, our expectation is that only the transverse states (λ =1, 2), are physical states. For the moment being we have ignored the µ gauge fixing condition phys|∂ Aµ|phys = 0, but its meaning is that only part of the total Hilbert space is physical. Therefore the relevant thing is to show that the states satisfying the Lorentz condition have positive norm. To discuss the gauge fixing condition, let us notice that formulated in the way we did, being bilinear in the states, it could destroy the linearity of the Hilbert space. So we will try to modify the condition in a linear one µ ∂ Aµ|phys. = 0 (1.181) But this would be a too strong requirement. Not even the vacuum state satisfies it. However, if we consider the positive and negative frequency parts of the field 3 3 (+) d k (λ) −ikx (−) (+) † Aµ (x)= µ (k)aλ(k)e ,Aµ (x)=(A (x)) (1.182) k 3 2ω (2π) λ=0

23 it is possible to weaken the condition, and require

µ (+) ∂ Aµ (x)|phys. = 0 (1.183)

This allows us to satisfy automatically the original requirement

µ (+) µ (−) phys.|(∂ Aµ + ∂ Aµ )|phys. = 0 (1.184)

(+) To make this condition more explicit let us evaluate the four divergence of Aµ 3 µ (+) d k −ikx µ (λ) i∂ Aµ (x)= e k µ (k)aλ(k) (1.185) k 3 2ω (2π) λ=0,3

Using eq. (1.167), we get

µ (3) µ (0) k µ = −(n · k),kµ =(n · k) (1.186) from which [a0(k) − a3(k)]|phys. = 0 (1.187) Notice that

− † − † − 3 − 3 − [a0(k) a3(k),a0(k ) a3(k )] = δ (k k )+δ (k k ) = 0 (1.188)

Let us denote by Φk(n0,n3)thestatewithn0 scalar photons (that is with polarization (0) (3) µ (k)), and with n3 longitudinal photons (that is with polarization µ (k)). Then the following states satisfy the condition (1.183)

(m) 1 † † m Φ = (a (k) − a (k)) Φ (0, 0) (1.189) k m! 0 3 k These states have vanishing norm

|| (m)||2 Φk = 0 (1.190) More generally we can make the following observation. Let us consider the number operator for scalar and longitudinal photons 3 † − † N = d k (a3(k)a3(k) a0(k)a0(k)) (1.191)

Notice the minus sign that is a consequence of the commutation relations, and it ensures that N has positive eigenvalues. For instance † | − 3 † † | † | Na0(k) 0 = d k a0(k )[a0(k ),a0(k)] 0 = a0(k) 0 (1.192)

24 Let us consider a physical state with a total number n of scalar and longitudinal photons. Then ϕn|N|ϕn = 0 (1.193) since a0 and a3 act in the same way on a physical state (see eq. (1.187)). It follows

n ϕn|ϕn = 0 (1.194)

Therefore all the physical states with a total deﬁnite number of scalar and longitudinal photons have zero norm, except for the vacuum state (n =0).Then

ϕn|ϕn = δn,0 (1.195)

A generic physical state with zero transverse photons is a linear superposition of the previous states |ϕ = c0|ϕ0 + ci|ϕi (1.196) i=0 This state has a positive deﬁnite norm

2 ϕ|ϕ = |c0| ≥ 0 (1.197)

The proof that a physical state has a positive norm can be extended to the case in which also transverse photons are present. Of course, the coeﬃcients ci, appearing in the expression of a physical state, are completely arbitrary, but this is not going to modify the values of the observables. For instance, consider the hamiltonian, we have 3 µ H = d x :[Π A˙ µ −L]: 3 µ0 λ 1 µν 1 λ 2 = d x : F A˙ µ − (∂ Aλ)A˙ + Fµν F + (∂ Aλ) : (1.198) 0 4 2

One can easily show the hamiltonian is given by the sum of all the degrees of freedom appearing in Aµ 3 1 3 2 2 2 2 H = d x : A˙ +(∇ Ai) − A˙ − ∇ A : 2 i 0 0 i=1 3 3 † − † = d kωk : aλ(k)aλ(k) a0(k)a0(k) : (1.199) λ=1

Since on the physical states a0 and a3 act in the same way, we get

2 3 † phys.|H|phys. = phys.| d kωk aλ(k)aλ(k)|phys. (1.200) λ=1

25 The generic physical state is of the form |ϕT ⊗|ϕ .with|ϕ deﬁned as in eq. (1.196). Since only |ϕT , contributes to the evaluation of an observable quantity , we can always choose |ϕ proportional to |ϕ0 . However, this does not mean that we are always working in the restricted physical space, because in a sum over the intermediate states we need to include all the degrees of freedom. This is crucial for the explicit covariance and locality of the theory. The arbitrariness in deﬁning the state |ϕ has, in fact, a very simple interpretation. It corresponds to add to Aµ a four gradient, that is it corresponds to perform a gauge transformation. Consider the following matrix element

ϕ|Aµ(x)|ϕ = cncm ϕn|Aµ(x)|ϕm (1.201) n,m

Since Aµ change the occupation number by one unit and all the states |ϕn have zero norm (except for the state with n = 0), the only non vanishing contributions come from n =0,m =1andn =1,m =0 d3k | | | −ikx (3) (0) | ϕ Aµ(x) ϕ = c0c1 0 e [µ (k)a3(k)+µ (k)a0(k)] ϕ1 +c.c. 2ωk(2π)3 (1.202) In order to satisfy the gauge condition the state |ϕ1 must be of the form | 3 † − † | ϕ1 = d qf(q)[a3(q) a0(q)] 0 (1.203) and therefore 3 | | d k (3) (0) −ikx ϕ Aµ(x) ϕ = [µ (k)+µ (k)][c0c1e f(k)+c.c.] (1.204) 2ωk(2π)3

From eqs. (1.167) and (1.168) we have

kµ (3) + (0) = (1.205) µ µ (k · n) from which ϕ|Aµ(x)|ϕ = ∂µΛ(x) (1.206) with 3 d k 1 −ikx Λ(x)= (ic0c1e f(k)+c.c.) (1.207) 2ωk(2π)3 n · k

It is important to notice that this gauge transformation leaves Aµ in the Lorentz gauge, since Λ = 0 (1.208)

26 because the momentum k inside the integral satisﬁes k2 = 0. From the expansion in eq. (1.172) one gets immediately the expression for the photon propagator − d4k e ikx 0|T (Aµ(x)Aν(y)|0 = −igµν (1.209) (2π)4 k2 + i

By deﬁning d4k − D(x)= e ikxD(k) (1.210) (2π)4 and 1 D(k)=− (1.211) k2 + i we get 0|T (Aµ(x)Aν (y)|0 = igµν D(x − y) (1.212)

1.5 Perturbation Theory

1.5.1 The Scattering Matrix In order to describe a scattering process we will assign to the vector of state a condition at t = −∞ |Φ(−∞) ≡|Φi (1.213) where the state Φi will be specified by assigning the set of incoming free particles in terms of eigenstates of momentum, spin and so on. For instance, in QED we will have to specify how many electrons, positrons and photons are in the initial state and we will have to specify their momenta, the spin projection of fermions and the polarization of the photons. The equations of motion will tell us how this state evolves with time and it will be possible to evaluate the state at t =+∞,where, ideally, we will detect the final states. In practice the preparation and the detection processes are made at some finite times. It follows that our ideal description will be correct only if these times are much bigger than the typical interaction time of the scattering process. Once we know Φ(+∞), we are interested to evaluate the probability amplitude of detecting at t =+∞ a given set of free particles specified by a vector state Φf . The amplitude is

Sfi = Φf |Φ(+∞) (1.214)

We will deﬁne the S matrix as the operator that give us |Φ(+∞) once we know |Φ(−∞) |Φ(+∞) = S|Φ(−∞) (1.215)

The amplitude Sfi is then Sfi = Φf |S|Φi (1.216)

27 Therefore Sfi is the S matrix element between free states (|Φi and Φf | are called in and out states respectively). In the interaction representation defined by iH0t iH0t −iH0t |Φ(t) = e S |ΦS(t) ,O(t)=e S OS(t)e S (1.217) 0 where the index S identifies the Schrödinger representation and HS is the free hamiltonian, the state vectors satisfy the Schrödinger equation with the interaction hamiltonian (evaluated in the interaction picture) ∂ i |Φ(t) = HI |Φ(t) (1.218) ∂t whereas the operators evolve with the free hamiltonian. To evaluate the S matrix we first transform the Schrödinger equation in the interaction representation in an integral equation t I |Φ(t) = |Φ(−∞) −i dt1 H (t1)|Φ(t1) (1.219) −∞ One can verify that this is indeed a solution, and that it satisfies explicitly the boundary condition at t = −∞. The perturbative expansion consists in evaluating |Φ(t) by iterating this integral equation. The result in terms of ordered T -products is ∞ n +∞ +∞ (−i) I I S =1+ dt1 ··· dtn T H (t1) ···H (tn) (1.220) n! −∞ −∞ n=1 The T -product of n terms means that the factors have to be written from left to right with decreasing times. For instance, if t ≥ t ≥···≥tn,then 1 2 I I I I T H (t1) ···H (tn) = H (t1) ···H (tn) (1.221) This result can be written in a more compact form by introducing the T -ordered exponential +∞ −i dt HI(t) S = T e −∞ (1.222) This expression is a symbolic one and it is really defined by its series expansion. The motivation for introducing the T -ordered exponential is that it satisfies the following factorization property t3 t3 t2 O(t)dt O(t)dt O(t)dt T e t1 = T e t2 T e t1 (1.223) If there are no derivative interactions we have +∞ − I 4 i dt H (t) +i d x Lint S = T e −∞ = T e (1.224) It follows that if the theory is Lorentz invariant, also the S matrix enjoys the same property.

28 1.5.2 Wick’s theorem The matrix elements of the S matrix between free particle states can be expressed as vacuum expectation values (VEV’s) of T -products. These VEVs satisfy an important theorem due to Wick that states that the T -products of an arbitrary number of free fields (the ones we have to do in the interaction representation) can be expressed as combinations of T -products among two fields, that is in terms of Feynman propagators. The Wick’s theorem is summarized in the following equation −i d4xj(x)φ(x) T e 1 −i d4xj(x)φ(x) − d4xd4yj(x)j(y) 0|T (φ(x)φ(y))|0 =: e : e 2 (1.225) where φ(x) is a free real scalar field and j(x) an ordinary real function. The previous formula can be easily extended to charged scalar, fermionic and photon fields. The Wick’s theorem is then obtained by expanding both sides of this equation in powers of j(x) and taking the VEV of both sides. Let us now expand both sides of eq. (1.225) in a series of j(x) and compare term by term. We will use the simplified notation φi ≡ φ(xi). We get T (φ)= :φ : (1.226)

T (φ1φ2)= :φ1φ2 :+ 0|T (φ1φ2)|0 (1.227)

3 T (φ1φ2φ3)= :φ1φ2φ3 :+ : φi : 0|T (φjφk)|0 (1.228) i= j= k=1

4 T (φ1φ2φ3φ4)= :φ1φ2φ3φ4 :+ : φiφj : 0|T (φkφl)|0 i= j= k= l=1 + 0|T (φiφj)|0 0|T (φkφl)|0 (1.229) and so on. By taking the VEV of these expression, and recalling that the VEV of a normal product is zero, we get the Wick’s theorem. The T -product of two field operators is sometimes called the contraction of the two operators. Therefore to evaluate the VEV of a T -product of an arbitrary number of free fields, it is enough to consider all the possible contractions of the fields appearing in the T -product. For instance, from the last of the previous relations we get

4 0|T (φ1φ2φ3φ4)|0 = 0|T (φiφj)|0 0|T (φkφl)|0 (1.230) i= j= k= l=1

29 An analogous theorem holds for the photon field. For the fermions one has to remember that the T -product is defined in a slightly different way. This gives a minus sign any time we have a permutation of the fermion fields which is odd with respect to the original ordering. As an illustration the previous formula becomes

4 0|T (ψ1ψ2ψ3ψ4)|0 = σP 0|T (ψiψj)|0 0|T (ψkψl)|0 (1.231) i= j= k= l=1 where σP = ±1 is the sign of the permutation (i, j, k, l) with respect to the fundamental one (1, 2, 3, 4) appearing on the right hand side. More explicitly

0|T (ψ1ψ2ψ3ψ4)|0 = 0|T (ψ1ψ2)|0 0|T (ψ3ψ4)|0

− 0|T (ψ1ψ3)|0 0|T (ψ2ψ4)|0

+ 0|T (ψ1ψ4)|0 0|T (ψ2ψ3)|0 (1.232)

1.5.3 Feynman diagrams in momentum space We will discuss here the Feynman rules for the scalar and spinor QED. Let us recall that the electromagnetic interaction is introduced via the minimal substitution

∂µ → ∂µ + ieAµ (1.233)

For instance, for the charged scalar ﬁeld we get

† µ 2 † 1 µν L = ∂µφ ∂ φ − m φ φ − Fµν F → free 4 † µ µ 2 † 1 µν → [(∂µ + ieAµ)φ)] [(∂ + ieA ) φ] − m φ φ − Fµν F (1.234) 4 The interaction part is given by † † µ 2 2 † Lint. = −ie φ ∂µφ − (∂µφ )φ A + e A φ φ (1.235)

Therefore the electromagnetic ﬁeld is coupled to the current † † jµ = ie φ ∂µφ − (∂µφ )φ (1.236) but another interaction term appears (the seagull term). In the spinor case we have a simpler situation

1 µν 1 µν L = ψ¯(i∂ˆ − m)ψ − Fµν F → ψ¯(i∂ˆ − eAˆ − m)ψ − Fµν F (1.237) free 4 4 giving the interaction term ¯ µ Lint = −eψγµψA (1.238)

30 In a typical experiment in particle physics we prepare beams of particles with definite momentum, polarization, etc. At the same time we measure momenta and polarizations of the final states. For thisreasonitisconvenienttoworkinamo- mentum representation. The result for a generic matrix element of the S matrix can be expressed in the following form 1 m f|S|i =(2π)4δ4( p − p ) M (1.239) in out 2EV VE in out ext. bosons ext. fermions Here f = i, and we are using the box normalization. The Feynman amplitude M (often one defines a matrix T wich differs from M by a factor i) can be obtained by drawing at a given perturbative order all the connected and topologically inequivalent Feynman’s diagrams. The diagrams are obtained by combining together the graphics elements given in Figures 1.5.1 and 1.5.2. The amplitude M is given by the sum of the amplitudes associated to these diagrams, after extraction of the factor (2π)4δ4( p) in (1.239). . Fermion line

Photon line

Scalar line

Fig. 1.5.1 - Graphical representation for both internal and external particle lines.

QED vertex

Scalar QED seagull

Fig. 1.5.2 - Graphical representation for the vertices in scalar and spinor QED.

31 At each diagram we associate an amplitude obtained by multiplying together the following factors

λ • for each ingoing and/or outgoing external photon line a factor µ (or its complex conjugate for outgoing photons if we consider complex polarizations as the circular one)

• for each ingoing and/or outgoing external fermionic line a factor u(p) and/or u¯(p)

• for each ingoing and/or outgoing external antifermionic line a factorv ¯(p) and/or v(p)

• for each internal scalar line a factor i∆F (p) (propagator)

• for each internal photon line a factor igµν D(p) (propagator)

• for each internal fermion line a factor iSF (p) (propagator) • − 4 4 − for each vertex in scalar QED a factor ie(p + p )(2π) δ ( in pi out pf ), where p and p are the momenta of the scalar particle (taking p ingoing and p outgoing) • 2 4 4 − For each seagull term in scalar QED a factor 2ie gµν (2π) δ ( in pi out pf ) • − 4 4 − for each vertex in spinor QED a factor ieγµ(2π) δ ( ingoing pi outgoing pf ) • Integrate over all the internal momenta with measure d4p/(2π)4.Thisgives rise automatically to the conservation of the total four-momentum.

• In a graph containing closed lines (loops), a factor (−1) for each fermionic loop (see Fig. 1.5.3)

Fermion loop

Fig. 1.5.3 - A fermion loop can appear inside any Feynman diagram (attaching the rest of the diagram by photonic lines).

In the following we will use need also the rules for the interaction with a static external e.m. ﬁeld. The correspending graphical element is given in Fig. 1.5.4. The

32 rule is to treat it as a normal photon, except that one needs to substitute the wave function of the photon / 1 1 2 (λ)(q) (1.240) 2EV µ with the Fourier transform of the external field ext 3 −iq · x ext Aµ (q )= d qe Aµ (x ) (1.241) where q is the momentum in transfer at the vertex where the external field line is attached to. Also, since the static external field breaks translational invariance, we 3 − loose the factor (2π) δ( in pin out pout). .

Fig. 1.5.4 - Graphical representation for an external electromagnetic ﬁeld. The upper end of the line can be attached to any fermion line.

1.5.4 The cross-section Let us consider a scattering process with a set of initial particles with four momenta pi =(Ei,pi) which collide and produce a set of ﬁnal particles with four momenta pf =(Ef ,pf ). From the rules of the previous Chapter we know that each external photon line contributes with a factor (1/2VE)1/2, whereas each external fermionic 1/2 line contributes with (m/V E) . Furthermore the conservation of the total four 4 4 momentum gives a term (2π) δ ( i pi − f pf ). If we exclude the uninteresting case f = i we can write / 1/2 1 2 4 4 m 1 Sfi =(2π) δ pi − pf M (1.242) VE 2VE i f fermioni bosoni where M is the Feynman amplitude which can be evaluated by using the rules of the previous Section. Let us consider the typical case of a two particle collision giving rise to an N particles ﬁnal state. Therefore, the probability per unit time of the transition is 2 given by w = |Sfi| with 4 4 1 1 2 w = V (2π) δ (Pf − Pi) (2m)|M| (1.243) 2VEi 2VEf i f fermioni

33 where, for reasons of convenience, we have written m 1 =(2m) (1.244) VE 2VE w gives the probability per unit time of a transition toward a state with well defined quantum numbers, but we are rather interested to the final states having momenta between pf and pf + dpf . Since in the volume V the momentum is given by p = 2πn/L, the number of final states is given by L 3 d3p (1.245) 2π

The cross-section is deﬁned as the probability per unit time divided by the ﬂux of the ingoing particles, and has the dimensions of a length to the square

[t−1 2t]=[ 2] (1.246)

In fact the ﬂux is deﬁned as vrelρ = vrel/V , since in our normalization we have a particle in the volume V ,andvrel is the relative velocity of the ingoing particles. For the bosons this follows from the normalization condition of the functions fk(x). For the fermions recall that in the box normalization the wave function is m − u(p)e ipx (1.247) EV from which m d3xρ(x)= d3x u†(p)u(p) = 1 (1.248) V V VE

Then the cross-section for getting the ﬁnal states with momenta between pf e pf +dpf is given by 3 V V Vd pf F dσ = w dN = w 3 (1.249) vrel vrel f (2π) We obtain

3 V Vd pf 4 4 1 1 2 f − i |M| dσ = 3 V (2π) δ (P P ) 2 (2m) vrel (2π) 4V E1E2 2VEf f f fermioni 3 4 4 1 d pf 2 f − i |M| =(2π) δ (P P ) (2m) 3 (1.250) 4E1E2vrel (2π) 2Ef fermioni f

Notice that the dependence on the quantization volume V disappears, as it should be, in the ﬁnal equation. Furthermore, the total cross-section, which is obtained by integrating the previous expression over all the ﬁnal momenta, is Lorentz invariant.

34 In fact, as it follows from the Feynman rules, M is invariant, as the factors d3p/2E. Furthermore p1 p2 vrel = v1 − v2 = − (1.251) E1 E2 but in the frame where the particle 2 is at rest (laboratory frame) we have p2 = (m2,0) and vrel = v1,fromwhich |p | | | 1 | | 2 − 2 E1E2 vrel = E1m2 = m2 p1 = m2 E1 m1 E1 2 2 − 2 2 · 2 − 2 2 = m2E1 m1m2 = (p1 p2) m1m2 (1.252)

We see that also this factor is Lorentz invariant.

35 Chapter 2

One-loop renormalization

2.1 Divergences of the Feynman integrals

Let us consider the Coulomb scattering. If we expand the S matrix up to the third order in the electric charge, and we assume that the external ﬁeld is weak enough such to take only the ﬁrst order, we can easily see that the relevant Feynman diagrams are the ones of Fig. 2.1.1 . .

p p' p p' = p'-p p'-p p p' p- k p p'- k p' k k p'-p p'-p p p' a) b) k p p' p-k p'-k

k k+q p'-p ' c)p -p d)

Fig. 2.1.1 - The Feynman diagram for the Coulomb scattering at the third order in the electric charge and at the ﬁrst order in the external ﬁeld.

36 The Coulomb scattering can be used, in principle, to deﬁne the physical electric charge of the electron. This is done assuming that the amplitude is linear in ephys, from which we get an expansion of the type

3 2 ephys = e + a2e + ···= e(1 + a2e + ···) (2.1) in terms of the parameter e which appears in the original lagrangian. The ﬁrst problem we encounter is that we would like to have the results of our calculation in terms of measured quantities as ephys. This could be done by inverting the previous expansion, but, and here comes the second problem, the coeﬃcient of the expansion are divergent quantities. To show this, consider, for instance the self-energy contribution to one of the external photons (as the one in Fig. 2.1.1a). We have µ ext i 2 Ma =¯u(p )(−ieγ )A (p − p) ie Σ(p) u(p) (2.2) µ pˆ − m + i where 4 − µν 2 d k ig i ie Σ(p)= (−ieγµ) (−ieγν) (2π)4 k2 + i − ˆ − pˆ k m + i 4 2 d k 1 1 µ = −e γµ γ (2.3) (2π)4 k2 + i pˆ − kˆ − m + i or 4 d k pˆ − kˆ + m µ 1 Σ(p)=i γµ γ (2.4) (2π)4 (p − k)2 − m2 + i k2 + i For large momentum, k, the integrand behaves as 1/k3 and the integral diverges linearly. Analogously one can check that all the other third order contributions diverge. Let us write explicitly the amplitudes for the other diagrams

2 i µ ext Mb =¯u(p )ie Σ(p ) (−ieγ )A (p − p)u(p) (2.5) pˆ − m + i µ − µ igµν 2 νρ ext Mc =¯u(p )(−ieγ ) ie Π (q)A (p − p)u(p),q= p − p (2.6) q2 + i ρ where d4k i i ie2Πµν (q)=(−1) Tr (−ieγµ) (−ieγν ) (2.7) (2π)4 kˆ +ˆq − m + i kˆ − m + i (the minus sign originates from the fermion loop) and therefore d4k 1 1 Πµν (q)=i Tr γµ γν (2.8) (2π)4 kˆ +ˆq − m kˆ − m + i The last contribution is

2 µ ext Md =¯u(p )(−ie)e Λ (p ,p)u(p)Aµ (p − p) (2.9)

37 where 4 d k i i −igαβ e2Λµ(p,p)= (−ieγα) γµ (−ieγβ) (2π)4 pˆ − kˆ − m + i pˆ − kˆ − m + i k2 + i (2.10) or 4 µ d k α 1 µ 1 1 Λ (p ,p)=−i γ γ γα (2.11) (2π)4 pˆ − kˆ − m + i pˆ − kˆ − m + i k2 + i The problem of the divergences is a serious one and in order to give some sense to the theory we have to define a way to define our integrals. This is what is called the regularization procedure of the Feynman integrals. That is we give a prescription in order to make the integrals finite. This can be done in various ways, as introducing an ultraviolet cut-off, or, as we shall see later by the more convenient means of dimensional regularization. However, we want that the theory does not depend on the way in which we define the integrals, otherwise we would have to look for some physical meaning of the regularization procedure we choose. This bring us to the other problem, the inversion of eq. (2.1). Since now the coefficients are finite we can indeed perform the inversion and obtaining e as a function of ephys and obtain all the observables in terms of the physical electric charge (that is the one measured in the Coulomb scattering). By doing so, a priori we will introduce in the observables a dependence on the renormalization procedure. We will say that the theory is renormalizable when this dependence cancels out. Thinking to the regularization in terms of a cut-off this means that considering the observable quantities in terms of ephys, and removing the cut-off (that is by taking the limit for the cut-off going to the infinity), the result should be finite. Of course, this cancellation is not obvious at all, and in fact in most of the theories this does not happen. However there is a restrict class of renormalizable theories, as for instance QED. We will not discuss the renormalization at all order and neither we will prove which criteria a theory should satisfy in order to be renormalizable. We will give these criteria without a proof but we will try only to justify them On a physical basis. As far QED is concerned we will study in detail the renormalization at one-loop.

2.2 Higher order corrections

In this section we will illustrate, in a simpliﬁed context, the previous considerations. Consider again the Coulomb scattering. At the lowest order the Feynman amplitude in presence of an external static electromagnetic ﬁeld is simply given by the following expression (see Fig. 2.1.1), (f = i)

m (1) Sfi =(2π)δ(Ef − Ei) M (2.12) VE where (1) µ ext M =¯u(p )(−ieγ )u(p)Aµ (p − p ) (2.13)

38 and ext 3 ext −iq · x Aµ (q)= d xAµ (x)e (2.14) is the Fourier transform of the static e.m. ﬁeld. In the case of a point-like source, of charge Ze we have Ze Aext(x)=(− ,0) (2.15) µ 4π|x| For the following it is more convenient to re-express the previous amplitude in terms of the external current generating the external e.m. ﬁeld. Since we have

Aµ = jµ (2.16) we can invert this equation in momentum space, by getting

gµν ν Aµ(q)=− j (q) (2.17) q2 + i from which − µν (1) ig ext M =¯u(p )(−ieγµ)u(p) (−ij (p − p)) (2.18) q2 + i ν with ext 3 jµ (x)=(Zeδ (x),0) (2.19) From these expressions one can get easily the diﬀerential cross-section dσ/dΩ. This quantity could be measured and then we could compare the result with the theoretical result obtained from the previous expression. In this way we get a measure of the electric charge. However the result is tied to a theoretical equation evaluated at the lowest order in e. In order to get a better determination one has to evaluate the higher order corrections, and again extract the value of e from the data. The lowest order result is at the order e2 in the electric charge. the next-to-leading contribu- tionsarattheordere4. For instance, the diagram in Fig. 2.1.1c gives one of these contribution. This is usually referred to as the vacuum polarization contribution and in this Section we will consider only this correction, just to exemplify how the renormalization procedure works. This contribution is given in eqs. (2.6) and (2.7). The total result is − µν − µρ − λν (1) ig ig 2 ig xt M + Mc =¯u(p )(−ieγµ)u(p) + (ie Πρλ) (−ij (q)) q2 + i q2 + i q2 + i ν (2.20) The sum gives a result very similar the to one obtained from M(1) alone, but with the photon propagator modiﬁed in the following way

− µν − µν − µρ − λν − µν − 2 µν ig ig ig 2 2 ig ig ie Π → + (ie Πρλ(q )) = + (2.21) q2 q2 q2 q2 q2 q4

39 As discussed in the previous Section Πµν is a divergent quantity. In fact, at high values of the momentum it behaves as d4p (2.22) p2 and therefore it diverges in a quadratic way. This follows by evaluating the integral by cutting the upper limit by a cut-oﬀ Λ. As we shall see later on, the real divergence is weaker due to the gauge invariance. We will see that the correct behaviour is given by d4p (2.23) p4 The corresponding divergence is only logarithmic, but in any case the integral is divergent. By evaluating it with a cut-oﬀ we get

2 2 2 Πµν (q )=−gµν q I(q )+··· (2.24)

The terms omitted are proportional to qµqν, and since the photon is always attached to a conserved current, they do not contribute to the ﬁnal result. The functions I(q2) can be evaluated by various tricks (we will evaluate it explicitly in the following) and the result is 1 Λ2 1 1 q2z(1 − z) 2 − − − I(q )= 2 log 2 2 dz z(1 z)log 1 2 (2.25) 12π m 2π 0 m For small values of q2 (−q2 << m2) one can expand the logarithm obtaining

1 Λ2 1 q2 I(q2) ≈ log + (2.26) 12π2 m2 60π2 m2 Summarizing the propagator undergoes the following modiﬁcation

−igλν −igµν → [1 − e2I(q2)] (2.27) q2 q2 This relation is shown in graphical terms in Fig. 2.2.1. . +

Fig. 2.2.1 - The vacuum polarization correction to the propagator.

The total result is then − 2 2 2 2 (1) i e Λ e q M + Mc =¯u(p )(−ieγ )u(p) 1 − log − (−iZe) (2.28) 0 q2 12π2 m2 60π2 m2

40 in the limit q → 0 the result looks exactly as at the lowest order, except that we need to introduce a renormalized charge eR given by

/ e2 Λ2 1 2 eR = e 1 − log (2.29) 12π2 m2

4 In term of eR and taking into account that we are working at the order e we can write − 2 2 (1) i eR q M + Mc =¯uf (ieRγ )ui 1 − (−iZeR) (2.30) 0 q2 60π2 m2 Now we compare this result with the experiment where we measure dσ/dΩinthe limit q → 0. We can extract again from the data the value of the electric charge, but this time we will measure eR. Notice that in this expression there are no more divergent quantities, since they have been eliminated by the definition of the renormalized charge. It must be stressed that the quantity that is fixed by the experiments is not the ”parameter” e appearing in the original lagrangian, but rather eR.This means that the quantity e is not really defined unless we specify how to relate it to measured quantities. This relation is usually referred to as the renormalization condition. As we have noticed the final expression we got does not contain any explicit divergent quantity. Furthermore, also eR, which is obtained from the experimental data, is a finite quantity. It follows that the parameter e,mustdivergefor Λ →∞. In other words, the divergence which we have obtained in the amplitude, is compensated by the divergence implicit in e. Therefore the procedure outlined here has meaning only if we start with an infinite charge. Said that, the renormalization procedure is a way of reorganizing the terms of the perturbative expansion in such a way that the parameter controlling it is the renormalized charge eR.Ifthisis possible, the various terms in the expansion turn out to be finite and the theory is said to be renormalizable. As already stressed this is not automatically satisfied. In fact renormalizability is not a generic feature of the relativistic quantum field theories. Rather the class of renormalizable theories (that is the theories for which the previous procedure is meaningful) is quite restricted. We have seen that the measured electric charge is different from the quantity appearing in the expression of the vertex. It will be convenient in the following to call this quantity eB (tha bare charge). On the contrary, the renormalized charge will be called e, rather than eR as done previously. For the moment being we have included only the vacuum polarization corrections, however if we restrict our problem to the charge renormalization, the divergent part coming from the self-energy and vertex corrections cancel out (see later). Let us consider now a process like e−µ− → e−µ−. Neglecting other corrections the total amplitude is given by Fig. 2.2.2. We will define the charge by choosing an arbitrary value of the square of the 2 2 2 momentum in transfer qµ, such that Q = −q = µ . In this way the renormalized charge depends on the scale µ. In the previous example we were using q → 0. The graphs in Fig. 2.2.2 sum up to give (omitting the contribution of the fermionic

41 . − e e − B e − e B e e B e B e B e B q + + e + e B e B B − µ e B e B − µ e B µ−

Fig. 2.2.2 - The vacuum polarization contributions to the e−µ− → e−µ− scattering. lines,) − µν − µρ − λν − µρ − λσ − τν 2 ig ig 2 ig ig 2 ig 2 ig eB + (ieBΠρλ) + (ieBΠρλ) (ieBΠστ ) +··· q2 q2 q2 q2 q2 q2 Q2=µ2 (2.31) By considering only the gµν contribution to Πµν ,weget − 2 igµν 2 2 4 2 2 e 1 − e I(q )+e I (q )+··· 2 2 (2.32) B q2 B B Q =µ This expression looks as the lowest order contribution to the propagator be deﬁning the renormalized electric charge as

2 2 2 2 4 2 2 e = eB[1 − eBI(µ )+eBI (µ )+···] (2.33)

Since the fermionic contribution is the same for all the diagrams, by calling it F (Q2), we get the amplitude in the form

2 2 2 2 4 2 2 M(eB)=eBF (Q )[1 − eBI(Q )+eBI (Q ) ···] (2.34)

2 2 2 2 where eBF (Q ) is the amplitude at the lowest order in eB.AsweknowI(Q )isa divergent quantity and therefore the amplitude is not deﬁned. However, following the discussion of the previous Section we can try to re-express everything in terms of the renormalized charge (2.33). We can easily invert by series the eq. (2.33) obtaining at the order e4

2 2 2 2 eB = e [1 + e I(µ )+···] (2.35)

42 By substituting this expression in M(e0)weget

2 2 2 2 2 2 M(eB)=e [1 + e I(µ )+···]F (Q )[1 − e I(Q )+···]= 2 2 2 2 2 = e F (Q )[1 − e (I(Q ) − I(µ )) + ···] ≡MR(e) (2.36)

From the expression (2.25) for I(Q2)weget 1 Λ2 1 1 Q2z(1 − z) I Q2 − dz z − z ( )= 2 log 2 2 (1 )log 1+ 2 (2.37) 12π m 2π 0 m Since the divergent part is momentum independent it cancels out in the expression for MR(e). Furthermore, the renormalized charge e will be measured through the cross-section, and therefore, by definition, it will be finite. Therefore, the charge renormalization makes the amplitude finite. Evaluating I(Q2) − I(µ2)perlargeQ2 (and µ2)weget 2α 1 m2 + Q2z(1 − z) e2(I(Q2) − I(µ2)) = − dz z(1 − z)log → π m2 + µ2z(1 − z) 0 2α 1 Q2 α Q2 →− dz z − z − (1 )log 2 = log 2 (2.38) π 0 µ 3π µ and in this limit 2 2 2 α Q MR(e)=e F (Q ) 1 − log + ··· (2.39) 3π µ2

At ﬁrst sight MR(e) depends on µ, but actually this is not the case. In fact, e is deﬁned through the eq. (2.33) and therefore it depends also on µ. What is going on is that the implicit dependecnce of e on µ cancels out the explicit dependence of MR(e). This follows from MR(e)=M(eB) and on the observation that M(eB) does not depend on µ. This can be formalized in a way that amounts to introduce the concept of renormalization group that we will discuss in much more detail later on. In fact, let us be more explicit by writing

M(eB)=MR(e(µ),µ) (2.40)

By diﬀerentiating this expression with respect to µ we get

dM(eB) ∂MR ∂MR ∂e = + = 0 (2.41) dµ ∂µ ∂e ∂µ This equation say formally that the amplitude expressed as a function of the renormalized charge is indeed independent on µ. This equation is the prototype of the renormalization group equations, and tells us how we have to change the coupling, when we change the deﬁnition scale µ. Going back to the original expression for M(e0), we see that this is given by 2 2 the tree amplitude eBF (Q ) times a geometric series arising from the iteration of

43 the vacuum polarization diagram. Therefore, it comes natural to deﬁne a coupling running with the momentum of the photon.

e2 e2 Q2 e2 − e2 I Q2 e4 I2 Q2 ··· B ( )= B[1 B ( )+ B ( )+ ]= 2 2 (2.42) 1+eBI(Q )

2 Also this expression is cooked up in terms of the divergent quantities eB and I(Q ), but again e2(Q2) is ﬁnite. This can be shown explicitly by evaluating the renormalized charge, e(µ2), in terms of the bare one

2 2 2 eB e (µ )= 2 2 (2.43) 1+eBI(µ ) We can put the last two equations in the form 1 1 1 1 I Q2 , I µ2 2 2 = 2 + ( ) 2 2 = 2 + ( ) (2.44) e (Q ) eB e (µ ) eB and subtracting them 1 1 − = I(Q2) − I(µ2) (2.45) e2(Q2) e2(µ2) we get e2(µ2) e2(Q2)= (2.46) 1+e2(µ2)[I(Q2) − I(µ2)]

By using this result we can write MR(e)as

2 2 2 MR(e)=e (Q )F (Q ) (2.47)

It follows that the true expansion parameter in the perturbative series is the running coupling e(Q2). By using eq. (2.38) we obtain

α(µ2) α(Q2)= (2.48) α(µ2) Q2 1 − log 3π µ2

We see that α(Q2) increases with Q2, therefore the perturbative approach looses validity for those values Q2 of such that α(Q2) ≈ 1, that is when

α(µ2) Q2 1 − log = α(µ2) (2.49) 3π µ2 or 1 2 3π − 1 Q 2 = e α(µ ) (2.50) µ2

44 For α(µ2) ≈ 1/137 we get Q2 ≈ e1281 ≈ 10556 (2.51) µ2 Actually α(Q2)isdivergentfor 3π 2 Q 2 = eα(µ ) ≈ e1291 ≈ 10560 (2.52) µ2

This value of Q2 is referred to as the Landau pole. Therefore in the case of QED the perturbative approach maintains its validity up to gigantic values of the momenta. This discussion shows clearly that the behaviour of the running coupling constant is determined by I(Q2), that is from the vaccum polarization contribution. In particular, α(Q2) increases with Q2 since in the expression e2(µ2)(I(Q2) − I(µ2)) the coeﬃcient of the logarithm is negative (equal to −α(µ2)/(3π)). We will see that in the so called non-abelian gauge theories the running is opposite, that is the theory becomes more and more perturbative, by increasing the energy.

2.3 The analysis by counterterms

The previous way of defining a renormalizable theory amounts to say that the original parameters in the lagrangian, as e, should be infinite and that their divergences should compensate the divergences of the Feynman diagrams. Then one can try to separate the infinite from the finite part of the parameters (this separation is ambigu- ous, see later). The infinite contributions are called counterterms, and by definition they have the same operator structure of the original terms in the lagrangian. On the other hand, the procedure of regularization can be performed by adding to the original lagrangian counterterms cooked in such a way that their contribution kills the divergent part of the Feynman integrals. This means that the coefficients of these counterterms have to be infinite. However they can also be regularized in the same way as the other integrals. We see that the theory will be renormalizable if the counterterms we add to make the theory finite have the same structure of the original terms in the lagrangian, in fact, if this is the case, they can be absorbed in the original parameters, which however are arbitrary, because they have to be fixed by the experiments (renormalization conditions). We will show now that at one-loop the divergent contributions come only from the three diagrams discussed before. In fact, let us define D as the superficial degree of divergence of a one-loop diagram the difference between 4 (the number of integration over the mimentum) and the number of powers of momentum coming from the propagators (we assume here that no powers of momenta are coming from the vertices, which is not true in scalar QED). We get D =4− IF − 2IB (2.53)

45 where IF and IB are the number of internal fermionic and bosonic lines. Going through the loop one has to have a number of vertices, V equal to the number of internal lines V = IF + IB (2.54)

Also, if at each vertex there are Ni lines of the particle of type i,thenwehave

NiV =2Ii + Ei (2.55) from which 2V =2IF + EF ,V=2IB + EB (2.56)

From these equations we can get V , IF and IB in terms of EF and EB 1 1 IF = EF + EB,IB = EF ,V= EF + EB (2.57) 2 2 from which 3 D =4− EF − EB (2.58) 2 It follows that the only one-loop superficially divergent diagrams in spinor QED are the ones in Fig. 2.3.1. In fact, all the diagrams with an odd number of external photons vanish. This is trivial at one-loop, since the trace of an odd number of γ matrices is zero. In general (Furry’s theorem) it follows from the conservation of charge conjugation, since the photon is an eigenstate of C with eigenvalue equal to -1. Furthermore, the effective degree of divergence can be less than the superficial degree. In fact, gauge invariance often lowers the divergence. For instance the diagram of Fig. 2.3.1 with four external photons (light-light scattering) has D =0 (logarithmic divergence), but gauge invariance makes it finite. Also the vacuum polarization diagram has an effective degree of divergence equal to 2. Therefore, there are only three divergent one-loop diagrams and all the divergences can be µ brought back to the three functions Σ(p), Πµν (q)eΛ(p ,p). This does not mean that an arbitrary diagram is not divergence, but it can be made finite if the previous functions are such. In such a case one has only to show that eliminating these three divergences (primitive divergences) the theory is automatically finite. In particular we will show that the divergent part of Σ(p) can be absorbed into the definition of the mass of the electron and a redefinition of the electron field (wave function renormalization). The divergence in Πµν , the photon self-energy, can be absorbed in the wave function renormalization of the photon (the mass of the photon is not renormalized due to the gauge invariance). And finally the divergence of Λµ(p ,p) goes into the definition of the parameter e. To realize this program we divide up the lagrangian density in two parts, one written in terms of the physical parameters, the other will contain the counterterms. We will call also the original parameters and fields of the theory the bare parameters and the bare fields and we will use an index B in order to distinguish them from the physical quantities. Therefore the

46 . E D E F B

0 2 2

0 4 0

2 0 1

2 1 0

Fig. 2.3.1 - The superﬁcially divergent one-loop diagrams in spinor QED two pieces of the lagrangian should look like as follows: the piece in terms of the physical parameters Lp µ 1 µν 1 µ 2 L = ψ¯(i∂ˆ − m)ψ − eψγ¯ µψA − Fµν F − (∂µA ) (2.59) p 4 2 and the counter terms piece Lc.t.

C E µ 2 L . . = iBψ¯∂ψˆ − Aψψ¯ − Fµν − (∂µA ) − eDψ¯Aψˆ (2.60) c t 4 2 and we have to require that la sum of these two contributions should coincide with the original lagrangian written in terms of the bare quantities. Adding together Lp and Lc.t. we get

µ 1+C µν L =(1+B)iψ¯∂ψˆ − (m + A)ψψ¯ − e(1 + D)ψγ¯ µψA − Fµν F + gauge − fixing 4 (2.61) where, for sake of simplicity, we have omitted the gauge fixing term. Defining the renormalization constant of the fields

Z1 =(1+D),Z2 =(1+B),Z3 =(1+C) (2.62) we write the bare ﬁelds as

1/2 µ 1/2 µ ψB = Z2 ψ, AB = Z3 A (2.63)

47 we obtain

m + A eZ1 µ 1 µν L = iψ¯B∂ψˆ B − ψ¯BψB − ψ¯BγµψBA − FB,µνF + gauge − ﬁxing Z 1/2 B 4 B 2 Z2Z3 (2.64) and putting m + A eZ1 mB = ,eB = (2.65) Z 1/2 2 Z2Z3 we get

µ 1 µν L = iψ¯B ∂ψˆ B − mBψ¯B ψB − eBψ¯BγµψBA − FB,µνF + gauge − fixing (2.66) B 4 B So we have succeeded in the wanted separation. Notice that the division of the parameters in physical and counter term part is well defined, because the finite piece is fixed to be an observable quantity. This requirement gives the renormalization conditions. The counter terms A, B, ... are determined recursively at each perturbative order in such a way to eliminate the divergent parts and to respect the renormalization conditions. We will see later how this works in practice at one-loop level. Another observation is that Z1 and Z2 have to do with the self-energy of the electron, and as such they depend on the electron mass. Therefore if we consider the theory for a different particle, as the muon, which has the same interactions as the electron and differs only for the value of the mass (mµ ≈ 200me), one would get a different bare electric charge for the two particles. Or, phrased in a different way, one would have to tune the bare electric charge at different values in order to get the same physical charge. This looks very unnatural, but the gauge invariance of the theory implies that at all the perturbative orders Z1 = Z2. As a consequence 1/2 eB = e/Z3 , and since Z3 comes from the photon self-energy, the relation between the bare and the physical electric charge is universal (that is it does not depend on the kind of charged particle under consideration). Summarizing, one starts dividing the bare lagrangian in two pieces. Then we regularize the theory giving some prescription to get finite Feynman integrals. The part containing the counter terms is determined, order by order, by requiring that the divergences of the Feynman integrals, which come about when removing the regularization, are cancelled out by the counter term contributions. Since the separation of an infinite quantity into an infinite plus a finite term is not well defined, we use the renormalization conditions, to fix the finite part. After evaluating a physical quantity we remove the regularization. Notice that although the counter terms are divergent quantity when we remove the cut off, we will order them according to the power of the coupling in which we are doing the perturbative calculation. That is we have a double limit, one in the coupling and the other in some parameter (regulator) which defines the regularization. The order of the limit is first to work at some order in the coupling, at fixed regulator, and then remove the regularization. Before going into the calculations for QED we want to illustrate some general results about the renormalization. If one considers only theory involving scalar,

48 fermion and massless spin 1 (as the photon) fields, it is not difficult to construct an algorithm which allows to evaluate the ultraviolet (that is for large momenta) divergence of any Feynman diagram. In the case of the electron self-energy (see Figs. 2.1.1a and 2.1.1b) one has an integration over the four momentum p and a behaviour of the integrand, coming from the propagators, as 1/p3, giving a linear divergence (it turns out that the divergence is only logarithmic). From this counting one can see that only the lagrangian densities containing monomials in the fields with mass dimension smaller or equal to the number of space-time dimensions have a finite number of divergent diagrams. It turns out also that these are renormalizable theories ( a part some small technicalities). The mass dimensions of the fields can be easily evaluated from the observation that the action is dimensionless in our units (/h = 1). Therefore, in n space-time dimensions, the lagrangian density, defined as dnx L (2.67) has a mass dimension n. Looking at the kinetic terms of the bosonic fields (two derivatives) and of the fermionic fields (one derivative), we see that n n − 1 dim[φ]=dim[Aµ]= − 1, dim[ψ]= (2.68) 2 2 In particular, in 4 dimensions the bosonic fields have dimension 1 and the fermionic 3/2. Then, we see that QED is renormalizable, since all the terms in the lagrangian density have dimensions smaller or equal to 4

µ µ 2 dim[ψψ¯ ]=3, dim[ψγ¯ µψA ]=4, dim[(∂ Aµ) ] = 4 (2.69)

The condition on the dimensions of the operators appearing in the lagrangian can be translated into a condition over the coupling constants. In fact each monomial Oi will appear multiplied by a coupling gi L = giOi (2.70) i therefore dim[gi]=4− dim[Oi] (2.71) The renormalizability requires dim[Oi] ≤ 4 (2.72) from which dim[gi] ≥ 0 (2.73) that is the couplings must have positive dimension in mass are to be dimensionless. In QED the only couplings are the mass of the electron and the electric charge

49 which is dimensionless. As a further example consider a single scalar ﬁeld. The most general renormalizable lagrangian density is characterized by two parameters

1 µ 1 2 2 3 4 L = ∂µφ∂ φ − m φ − ρφ − λφ (2.74) 2 2 Here ρ has dimension 1 and λ is dimensionless. We see that the linear σ-models are renormalizable theories. Giving these facts let us try to understand what makes renormalizable and non renormalizable theories different. In the renormalizable case, if we have written the most general lagrangian, the only divergent diagrams which appear are the ones corresponding to the processes described by the operators appearing in the lagrangian. Therefore adding to L the counter term Lc.t. = δgiOi (2.75) i we can choose the δgi in such a way to cancel, order by order, the divergences. The theory depends on a finite number of arbitrary parameters equal to the number of parameters gi. Therefore the theory is a predictive one. In the non renormalizable case, the number of divergent diagrams increase with the perturbative order. At each order we have to introduce new counter terms having an operator structure different from the original one. At the end the theory will depend on an infinite number of arbitrary parameters. As an example consider a fermionic theory with an interaction of the type (ψψ¯ )2. Since this term has dimension 6, the relative coupling has dimension -2 ¯ 2 Lint = −g2(ψψ) (2.76) At one loop the theory gives rise to the divergent diagrams of Fig. 2.3.2. The divergence of the first diagram can be absorbed into a counter term of the original type ¯ 2 −δg2(ψψ) (2.77) The other two need counter terms of the type

¯ 3 ¯ 4 δg3(ψψ) + δg4(ψψ) (2.78)

These counter terms originate new one-loop divergent diagrams, as for instance the ones in Fig. 2.3.3. The ﬁrst diagram modiﬁes the already introduced counter term (ψψ¯ )4, but the second one needs a new counter term

¯ 5 δg5(ψψ) (2.79)

This process never ends. The renormalization requirement restricts in a fantastic way the possible ﬁeld theories. However one could think that this requirement is too technical and one could imagine other ways of giving a meaning to lagrangians which do not satisfy this

50 .

a) b)

Fig. 2.3.2 - Divergent diagrams coming from the interaction (ψψ¯ )2. condition. But suppose we try to give a meaning to a non renormalizable lagrangian, simply requiring that it gives rise to a consistent theory at any energy. We will show that this does not happen. Consider again a theory with a four-fermion interaction. Since dim g2 = −2, if we consider the scattering ψ + ψ → ψ + ψ in the high energy limit (where we can neglect all the masses), on dimensional ground we get that the total cross-section behaves like ≈ 2 2 σ g2E (2.80) Analogously, in any non renormalizable theory, being there couplings with negative dimensions, the cross-section will increase with the energy. But the cross-section has to do with the S matrix which is unitary. Since a unitary matrix has eigenvalues of .

a) b)

Fig. 2.3.3 - Divergent diagrams coming from the interactions (ψψ¯ )2 and (ψψ¯ )4.

51 modulus 1, it follows that its matrix elements must be bounded. Translating this argument in the cross-section one gets the bound

c2 σ ≤ (2.81) E2 where c is some constant. For the previous example we get

2 g2E ≤ c (2.82)

This implies a violation of the S matrix unitarity at energies such that c E ≥ (2.83) g2 It follows that we can give a meaning also to non renormalizable theories, but only for a limited range of values of the energy. This range is ﬁxed by the value of the non renormalizable coupling. It is not diﬃculty to realize that non renormalizability and bad behaviour of the amplitudes at high energies are strictly connected.

2.4 Dimensional regularization of the Feynman integrals

As we have discussed in the previous Section we need a procedure to give sense at the otherwise divergent Feynman diagrams. The simplest of these procedures is just ∞ → Λ to introduce a cut-off Λ and define our integrals as 0 limΛ→∞ 0 ). Of course, in the spirit of renormalization we have first to perform the perturbative expansion and then take the limit over the cut-off. Although this procedure is very simple it results to be inadequate in situations like gauge theories. In fact one can show that the cut-off regularization breaks the translational invariance and creates problems in gauge theories. One kind of regularization which nowadays is very much used is the dimensional regularization. This consists in considering the integration in an arbitrary number of space-time dimensions and taking the limit of four dimensions at the end. This way of regularizing is very convenient because it respects all the symmetries. In fact, except for very few cases the symmetries do not depend on the number of space-time dimensions. Let us what is dimensional regularization about. We want to evaluate integrals of the type 4 I4(k)= d pF(p, k) (2.84) with F (p, k) ≈ p−2 or p−4. The idea is that integrating on a lower number of dimensions the integral improves the convergence properties in the ultraviolet. For

52 instance, if F (p, k) ≈ p−4, the integral is convergent in 2 and in 3 dimensions. Therefore, we would like to introduce a quantity I(ω,k)= d2ωpF(p, k) (2.85) to be regarded as a function of the complex variable ω. Then, if we can define a complex function I(ω,k) on the entire complex plane, with definite singularity, and as such that it coincides with I on some common domain, then by analytic continuation I and I define the same analytic function. A simple example of this procedure is given by the Euler’s Γ. This complex function is defined for Re z > 0 by the integral representation ∞ − z− Γ(z)= dt e tt 1 (2.86) 0 If Re z ≤ 0, the integral diverges as dt (2.87) t1+|Re z| in the limit t → 0.However it is easy to get a representation valid also for Re z ≤ 0. Let us divide the integration region in two parts defined by a parameter α α ∞ − z− − z− Γ(z)= dt e tt 1 + dt e tt 1 (2.88) 0 α Expanding the exponential in the first integral and integrating term by term we get ∞ n α ∞ (−1) n z− − z− Γ(z)= dt t + 1 + dt e tt 1 n! α n=0 0 ∞ n n+z ∞ (−1) α − z− = + dt e tt 1 (2.89) n! n + z α n=0 The second integral converges for any z since α>0. This expression coincides with the representation for the Γ function for Re z > 0, but it is defined also for Re z < 0 where it has simple poles located at z = −n. Therefore it is a meaningful expression on all the complex plane z. Notice that in order to isolate the divergences we need to introduce an arbitrary parameter α. However the result does not depend on the particular value of this parameter. This the Weierstrass representation of the Euler Γ(z). From this example we see that we need the following three steps • Find a domain where I(ω,k) is convergent. Typically this will be for Re ω<2. • Construct an analytic function identical to I(ω,k) in the domain of convergence, but defined on a larger domain including the point ω =2. • At the end of the calculation take the limit ω → 2.

53 2.5 Integration in arbitrary dimensions

Let us consider the integral N 2 IN = d pF (p ) (2.90) with N an integer number and p a vector in an Euclidean N-dimensional space. Since the integrand is invariant under rotations of the N-dimensional vector p,we can perform the angular integration by means of

N N−1 d p = dΩN p dp (2.91) where dΩN is the solid angle element in N-dimensions. Therefore dΩN = SN ,with SN the surface of the unit sphere in N-dimensions. Then

∞ N−1 2 IN = SN p F (p )dp (2.92) 0 The value of the sphere surface can be evaluated by the following trick. Consider

∞ + − 2 √ I = e x dx = π (2.93) −∞

By taking N of these factors we get 2 2 N −(x + ···+ xN ) N/2 I = dx1 ···dxN e 1 = π (2.94)

The same integral can be evaluated in polar coordinates

∞ 2 N/2 N−1 −ρ π = SN ρ e dρ (2.95) 0 By putting x = ρ2 we have

∞ N/2 1 N/2−1 −x 1 N π = SN x e dx = SN Γ (2.96) 2 0 2 2 where we have used the representation of the Euler Γ function given in the previous Section. Therefore N/2 2π SN = N (2.97) Γ 2 and N/2 ∞ π N/2−1 IN = N x F (x)dx (2.98) Γ 2 0 with x = p2.

54 The integrals we will be interested in are of the type N M d p I( ) = (2.99) N (p2 − a2 + i)A with p a vector in a Ndimensional Minkowski space. We can perform an anti- 0 clockwise rotation of 90 (Wick’s rotation) in the complex plane of p0 without hitting → any singularity . Then we do a change of variables p0 ip0, and we use the deﬁnition 2 →− 2 N 2 p p = i=1 pi , obtaining N (M) d p A I = i = i(−1) IN (2.100) N (−p2 − a2)A where IN is an integral of the kind discussed at the beginning of this Section with F (x)givenby F (x)=(x + a2)−A (2.101) It follows πN/2 ∞ xN/2−1 IN dx = N 2 A (2.102) Γ 2 0 (x + a ) By putting x = a2y we get N/2 ∞ 2 N/2−A π N/2−1 −A IN =(a ) N y (1 + y) dx (2.103) Γ 2 0 and recalling the integral representation for the Euler B(x, y) function (valid for Rex, y > 0) Γ(x)Γ(y) ∞ B(x, y)= = tx−1(1 + t)−(x+y)dt (2.104) Γ(x + y) 0 it follows − N/2 Γ(A N/2) 1 IN = π (2.105) Γ(A) (a2)A−N/2 We have obtained this representation for N/2 > 0andRe(A − N/2) > 0. But we know how to extend the Euler gamma-function to the entire complex plane, and therefore we can extend this formula to complex dimensions N =2ω

ω Γ(A − ω) 1 I ω = π (2.106) 2 Γ(A) (a2)A−ω

This shows that I2ω has simple poles located at ω = A, A +1, ··· (2.107) Therefore our integral will be perfectly deﬁned at all ω such that ω = A, A +1, ···. At the end we will have to consider the limit ω → 2. The original integral in Minkowski space is then d2ωp Γ(A − ω) 1 = iπω(−1)A (2.108) (p2 − a2)A Γ(A) (a2)A−ω

55 For the following it will be useful to derive another formula. Let us put in the previous equation p = p + k and b2 = −a2 + k2,then d2ωp Γ(A − ω) 1 = iπω(−1)A (2.109) (p2 +2p · k + k2 − a2)A Γ(A) (a2)A−ω from which d2ωp Γ(A − ω) 1 = iπω(−1)A (2.110) (p2 +2p · k + b2)A Γ(A) (k2 − b2)A−ω

Diﬀerentiating with respect to kµ we get various useful relations as pµ Γ(A − ω) −kµ d2ωp = iπω(−1)A (2.111) (p2 +2p · k + b2)A Γ(A) (k2 − b2)A−ω and ω A pµpν π (−1) d2ωp = i 2 · 2 A 2 − 2 A−ω (p +2p k + b ) Γ(A)(k b ) 1 2 2 × Γ(A − ω)kµkν − gµν (k − b )Γ(A − ω − 1) (2.112) 2 Since at the end of our calculation we will have to take the limit ω → 2, it will be useful to recall the expansion of the Gamma function around its poles 1 Γ()= − γ + O() (2.113) where γ =0.5772... (2.114) is the Euler-Mascheroni constant, and for (n ≥ 1): (−1)n 1 Γ(−n + )= + ψ(n +1)+O() (2.115) n! where 1 1 ψ(n +1)=1+ + ···+ − γ (2.116) 2 n 2.6 One loop regularization of QED

In this Section we will regularize, by using dimensional regularization, the relevant µ divergent quantities in QED, that is Σ(p), Πµν (q)eΛ (p, p ). Furthermore, in order to deﬁne the counter terms we will determine the expressions which become divergent in the limit of ω → 2. It will be also convenient to introduce a parameter µ such that these quantities have the same dimensions in the space with d =2ω as in d =4.

56 The algebra of the Dirac matrices is also easily extended to arbitrary dimensions d. For instance, starting from

[γµ,γν]+ =2gµν (2.117) we get µ γ γµ = d (2.118) and µ γµγνγ =(2− d)γν (2.119) Other relations can be obtained by starting from the algebraic properties of the γ-matrices. Let us start with the electron self-energy which we will require to have dimension 1 as in d = 4. From eq. (2.4) we have 2ω − ˆ 4−2ω d k pˆ k + m µ 1 Σ(p)=iµ γµ γ (2.120) (2π)2ω (p − k)2 − m2 + i k2 + i

In order to use the equations of the previous Section it is convenient to combine together the denominators of this expression into a single one. This is done by using a formula due to Feynman 1 1 dz = (2.121) − 2 ab 0 [az + b(1 z)] whichisprovenusing

b 1 1 1 − 1 1 dx = = 2 (2.122) ab b − a a b b − a a x and doing the change of variables

x = az + b(1 − z) (2.123)

We get 1 2ω − ˆ 4−2ω d k pˆ k + m µ Σ(p)=iµ dz γµ γ (2.124) 2ω − 2 − 2 2 − 2 0 (2π) [(p k) z m z + k (1 z)] The denominator can be written in the following way

[...]=p2z − m2z + k2 − 2p · kz (2.125) and the term p · k can be eliminated through the following change of variables k = k + pz. We ﬁnd

[...]=(p2 − m2)z +(k + pz)2 − 2p · (k + pz)z = k2 − m2z + p2z(1 − z) (2.126)

57 It follows (we put k = k) 1 2ω − − ˆ 4−2ω d k pˆ(1 z) k + m µ Σ(p)=iµ dz γµ γ (2.127) 2ω 2 − 2 2 − 2 0 (2π) [k m z + p z(1 z)] The linear term in k has vanishing integral, therefore 1 2ω − 4−2ω d k pˆ(1 z)+m µ Σ(p)=iµ dz γµ γ (2.128) 2ω 2 − 2 2 − 2 0 (2π) [k m z + p z(1 z)] and integrating over k 1 − − 4−2ω 1 ω Γ(2 ω) pˆ(1 z)+m µ Σ(p)=iµ dz iπ γµ γ (2.129) 2ω 2 − 2 − 2−ω 0 (2π) Γ(2) [m z p z(1 z)] By defining =4− 2ω we get 1 − 1 (4−)/2 pˆ(1 z)+m µ Σ(p)=−µ dz π Γ(/2)γµ γ (2.130) 4− 2 − 2 − /2 0 (2π) [m z p z(1 z)] Contracting the γ matrices 1 1 ( − 2)ˆp(1 − z)+(4− )m Σ(p)=− Γ(/2) dz(4πµ2)/2 (2.131) 2 2 − 2 − /2 16π 0 [m z p z(1 z)] we obtain 1 p / Σ( )= 2 Γ( 2) 16π 1 2 − 2 − −/2 × − − − − − m z p z(1 z) dz[2ˆp(1 z) 4m (ˆp(1 z) m)] 2 (2.132) 0 4πµ Defining m2z − p2z(1 − z) A =2ˆp(1 − z) − 4m, B = −pˆ(1 − z)+m, C = (2.133) 4πµ2 and expanding for → 0 1 1 2A Σ(p)= dz +2B − γA 1 − log C 16π2 2 0 1 1 2A − − = 2 dz A log C +2B γA 16π 0 1 1 = (ˆp − 4m) − [ˆp − 2m + γ(ˆp − 4m)] 8π2 16π2 1 2 − 2 − − 1 − − m z p z(1 z) 2 dz[ˆp(1 z) 2m]log 2 8π 0 4πµ 1 = (ˆp − 4m) + finite terms (2.134) 8π2

58 Consider now the vacuum polarization (see eq. (2.8)) d2ωk 1 1 Πµν (q)=iµ4−2ω Tr γµ γν (2π)2ω kˆ +ˆq − m kˆ − m 2ω d k Tr[γµ(kˆ + m)γν(kˆ +ˆq + m)] = iµ4−2ω (2.135) (2π)2ω (k2 − m2)((k + q)2 − m2) Using again the Feynman representation for the denominators 1 2ω d k Tr[γµ(kˆ + m)γν(kˆ +ˆq + m)] Πµν (q)=iµ4−2ω dz (2.136) 2ω 2 − 2 − 2 − 2 2 0 (2π) [(k m )(1 z)+((k + q) m )z] We write the denominator as [...]=k2 + q2z − m2 +2k · qz (2.137) through the change of variable k = k − qz we cancel the mixed term obtaining [...]=(k − qz)2 +2(k − qz) · qz + q2z − m2 = k2 + q2z(1 − z) − m2 (2.138) and therefore (we put again k = k) 1 2ω d k Tr[γµ(kˆ − qzˆ + m)γν(kˆ +ˆq(1 − z)+m)] Πµν (q)=iµ4−2ω dz 2ω 2 2 − − 2 2 0 (2π) [k + q z(1 z) m ] (2.139) Since the integral of the odd terms in k is zero, it is enough to evaluate the contribution of the even term to the trace ˆ ˆ 2 Tr[...]even = Tr[γµ(k − qzˆ + m)γν(k +ˆq(1 − z)]pari + m Tr[γµγν] 2 = Tr[γµkγˆ νkˆ] − Tr[γµqγˆ νqˆ]z(1 − z)+m Tr[γµγν] (2.140) If we deﬁne the γ as matrices of dimension 2ω × 2ω we can repeat the standard calculation by obtaining a factor 2ω instead of 4. Therefore

ω 2 2 2 Tr[...]even =2 [2kµkν − gµν k − (2qµqν − gµν q )z(1 − z)+m gµν ] ω 2 2 2 2 =2 [2kµkν − 2z(1 − z)(qµqν − gµν q ) − gµν (k − m + q z(1 − z)](2.141) and we ﬁnd 1 2ω d k 2kµkν Πµν (q)=iµ4−2ω2ω dz 2ω 2 2 − − 2 2 0 (2π) [k + q z(1 z) m ] 2 2z(1 − z)(qµqν − gµν q ) gµν − − (2.142) [k2 + q2z(1 − z) − m2]2 [k2 + q2z(1 − z) − m2] From the relations of the previous Section we get − 2ω 2pµpν ω Γ(1 ω) d p = −igµν π (2.143) [p2 − a2]2 (a2)1−ω

59 whereas 1 Γ(1 − ω) d2ωp = −iπω (2.144) [p2 − a2] (a2)1−ω Therefore the ﬁrst and the third contribution to the vacuum polarization cancel out and we are left with 1 2ω µν 4−2ω ω 2 d k 1 Π (q)=−iµ 2 (qµqν −gµν q ) dz 2z(1−z) 2ω 2 2 − − 2 2 0 (2π) [k + q z(1 z) m ] (2.145) Notice that the original integral was quadratically divergent, but due to the previous cancellation the divergence is only logarithmic. The reason is again gauge invari- µ ance. In fact it is possible to show that this implies q Πµν (q) = 0. Performing the momentum integration we have 1 ω − µν 4−2ω ω 2 iπ Γ(2 ω) Π (q)=−i2µ 2 (qµqν − gµν q ) dz z(1 − z) 2ω 2 − 2 − 2−ω 0 (2π) [m q z(1 z)] (2.146) By putting again =4− 2ω and expanding the previous expression 1 2−/2 µν 2−/2 2 π Γ(/2) Π (q)=2µ 2 (qµqν − gµν q ) dz z(1 − z) (2π)4− [m2 − q2z(1 − z)]/2 0 1 2 − 2 − −/2 2 2−/2 2 m q z(1 z) qµqν − gµν q dz z − z / = 2 2 ( ) (1 )Γ( 2) 2 16π 0 4πµ 2 2 = (4 − 2 log 2)(qµqν − gµν q ) 16π2 1 2 × dz z(1 − z)( − γ) 1 − log C (2.147) 0 2 where C is deﬁnite in eq. (2.133). Finally 1 2 8 Πµν (q)= (qµqν − gµν q ) dz z(1 − z) − 4γ − 4log2 1 − log C 8π2 2

1 2 = (qµqν − gµν q ) 2π2 1 γ m2 − q2z(1 − z) × − − dz z(1 − z)log (2.148) 3 6 2πµ2 or, in abbreviated way

1 2 1 Πµν (q)= (qµqν − gµν q ) + ﬁnite terms (2.149) 6π2 We have now to evaluate Λµ(p ,p). From eq. (2.11) we have 2ω − ˆ − ˆ µ 4−2ω d k α pˆ k + m µ pˆ k + m 1 Λ (p ,p)=−iµ γ γ γα (2.150) (2π)2ω (p − k)2 − m2 (p − k)2 − m2 k2

60 the general formula to reduce n denominators to a single one is n 1 n n 1 δ(1 − βi) − i=1 =(n 1)! dβi n n (2.151) ai [ βiai] i=1 0 i=1 i=1 To show this equation notice that n − n ∞ n αiai 1 i = dαie =1 (2.152) ai i=1 0 i=1

Introducing the identity ∞ n 1= dλδ(λ − αi) (2.153) 0 i=1 and changing variables αi = λβi n − n ∞ n n αiai 1 i = dαidλδ(λ − αi)e =1 ai i=1 0 i=1 i=1 n − ∞ n λ βiai n n dλ i = dβiλ e =1 δ(1 − βi) (2.154) λ 0 i=1 i=1

The integration over βi can be restricted to the interval [0, 1] due to the delta function, and furthermore ∞ − n−1 −ρλ (n 1)! dλλ e = n (2.155) 0 ρ In our case we get d2ωk 1 1−x µ p,p − iµ4−2ω dx dy Λ ( )= 2 2ω (2π) 0 0 α µ γ (ˆp − kˆ + m)γ (ˆp − kˆ + mγα × (2.156) [k2(1 − x − y)+(p − k)2x − m2x +(p − k)2y − m2y]3 The denominator is

[...]=k2 − m2(x + y)+p2x + p2y − 2k · (px + py) (2.157)

Changing variable, k = k + px + py

[...]=(k + px + py)2 − m2(x + y)+p2x + p2y − 2(k + px + py) · (px + py) = k 2 − m2(x + y)+p2x(1 − x)+p2y(1 − y) − 2p · pxy (2.158)

61 Letting again k → k 1 1−x d2ωk µ − 4−2ω Λ (p ,p)= 2iµ dx dy 2ω 0 0 (2π) α µ γ (ˆp (1 − y) − pxˆ − kˆ + m)γ (ˆp(1 − x) − pˆy − kˆ + m)γα × (2.159) [k2 − m2(x + y)+p2x(1 − x)+p2y(1 − y) − 2p · pxy]3 The odd term in k is zero after integration, the term in k2 is logarithmically divergent, whereas the remaining part is convergent. Separating the divergent piece, (1) (3) Λµ , from the convergent one, Λµ ,

(1) (2) Λµ =Λµ +Λµ (2.160) we get for the ﬁrst term 1 1−x d2ωk (1) p,p − iµ4−2ω dx dy Λµ ( )= 2 2ω 0 0 (2π) λ σ α k k γ γλγµγσγα [k2 − m2(x + y)+p2x(1 − x)+p2y(1 − y) − 2p · pxy]3 1 1−x iπω(−1)3 1 = −2iµ4−2ω dx dy − Γ(2 − ω) 0 0 Γ(3) 2 α µ λ γ γλγ γ γα [m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy]2−ω ω 1 1 1 1−x = µ4−2ω Γ(2 − ω) dx dy 2 4π 0 0 α µ λ γ γλγ γ γα (2.161) [m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy]2−ω Using the relation α µ λ 2 µ γ γλγ γ γα =(2− d) γ (2.162) we obtain ( =4− 2ω) 2−/2 1 1−x (1) 1 1 2 Λµ (p ,p)= µ Γ(/2)( − 2) γµ dx dy 2 4π 0 0 1 [m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy]/2 1 2 1 1−x − − µ = 2 γ [4 4]γ dx dy 32π 0 0 −/2 m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy 4πµ2 1 1 1 1 1−x µ − µ − µ = 2 γ 2 (γ +2)γ 2 γ dx dy 8π 16π 8π 0 0

62 m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy × log (2.163) 4πµ2 and ﬁnally (1) 1 Λ (p ,p)= γµ + ﬁnite terms (2.164) µ 8π2 In the convergent part we can put directly ω =2 i 1 1−x iπ2(−1)3 (2) p,p − dx dy Λµ ( )= 4 8π 0 0 Γ(3) α µ γ (ˆp (1 − y) − pxˆ + m)γ (ˆp(1 − x) − pˆ y + m)γα m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy 1 1 1−x − dx dy = 2 16π 0 0 α µ γ (ˆp (1 − y) − pxˆ + m)γ (ˆp(1 − x) − pˆ y + m)γα (2.165) m2(x + y) − p2x(1 − x) − p2y(1 − y)+2p · pxy

2.7 One loop renormalization

We summarize here the results of the previous Section 1 Σ(p)= (ˆp − 4m)+Σf (p) (2.166) 8π2 2 1 f 2 2 Πµν (q)=(qµqν − gµν q ) +Π (q) ≡ (qµqν − gµν q )Π(q ) (2.167) 6π2

1 f Λµ(p ,p)= γµ +Λ (p ,p) (2.168) 8π2 µ where the functions with the superscript f represent the ﬁnite contributions. Let us start discussing the electron self-energy. As shown in eq. (2.2), the eﬀect of Σ(p) is to correct the electron propagator. In fact we have (see Fig. 2.7.1):

i i 2 i SF (p)= + ie Σ(p) + ··· (2.169) pˆ − m pˆ − m pˆ − m

Fig. 2.7.1 - The loop expansion for the electron propagator. from which, at the same order in the perturbative expansion

− i e2Σ(p) 1 i SF (p)= 1+ = (2.170) pˆ − m pˆ − m pˆ − m + e2Σ(p)

63 Therefore the effect of the divergent terms is to modify the coefficients ofp ˆ and m: e2 e2 iS−1(p)=ˆp − m + e2Σ(p)=ˆp 1+ − m 1+ + finite terms (2.171) F 8π2 2π2

This allows us to deﬁne the counter terms to be added to the lagrangian expressed in terms of the physical parameters in such a way to cancel these divergences ¯ˆ ¯ (L1)ct = iBψ∂ψ − Aψψ (2.172)

(L1)ct modiﬁes the Feynman rules adding two particles interacting terms. These can be easily evaluated noticing that the expression of the propagator, taking into account (L1)ct,is i i Bpˆ − A ≈ 1 − (1 + B)ˆp − (m + A) pˆ − m pˆ − m i i i ≈ + (iBpˆ − iA) (2.173) pˆ − m pˆ − m pˆ − m where, consistently with our expansion we have taken only the ﬁrst order terms in A and B. We can associate to these two terms the diagrams of Fig. 2.7.2, with contributions −iA to the mass term, and iBpˆ top ˆ. .

-iA iBp/

Fig. 2.7.2 - The counter terms for the self-energy (/p in the ﬁgure should be read as pˆ).

The propagator at the second order in the coupling constant is then given by adding the diagrams of Fig. 2.7.3. At this order we get i i 2 i SF (p)= + ie Σ(p)+iBpˆ − iA (2.174) pˆ − m pˆ − m pˆ − m and the correction to the free propagator is given by e2 e2 e2Σ(p)+Bpˆ − A = + B pˆ − m + A + ﬁnite terms (2.175) 8π2 2π2

We can now ﬁx the counter terms by choosing e2 1 m B = − + F (2.176) 8π2 2 µ

64 .

Fig. 2.7.3 - The second order contributions at the electron propagator.

me2 1 m A = − + Fm (2.177) 2π2 µ with F2 andFm ﬁnite for → 0. Notice also that these two functions are dimensionless and arbitrary up to this moment. However they can be determined by the renormalization conditions, that is by ﬁxing the arbitrary constants appearing in the lagrangian. In fact, given

−1 (2) 2 iSF (p) ≡ Γ (p)=ˆp − m + Bpˆ − A + e Σ(p) (2.178) we can require that at the physical pole,p ˆ = m, the propagator coincides with the free propagator i SF (p) ≈ , forp ˆ = m (2.179) pˆ − m From here we get two conditions. The ﬁrst one is 2 2 2 (2) e 2 f e 1 me 1 0=Γ (ˆp = m)= (ˆp−4m)+e Σ (ˆp = m)− + F pˆ+ + Fm 8π2 8π2 2 2π2 (2.180) from which 2 2 2 f me me e Σ (ˆp = m) − F + Fm = 0 (2.181) 8π2 2 2π2 The second condition is ∂Γ(2)(p) = 1 (2.182) ∂pˆ pˆ=m

65 which gives f 2 2 ∂Σ (p) e e − F2 = 0 (2.183) ∂pˆ pˆ=m 8π2 One should be careful because these particular conditions of renormalization gives some problems related to the zero mass of the photon. In fact one finds some ill- defined integral in the infrared region. However these are harmless divergences, because giving a small to the photon and letting it to zero at the end of the calculations gives rise to finite results. Notice that these conditions of renormalization have the advantage of being expressed directly in terms of the measured parameters, as the electron mass. However, one could renormalize at an arbitrary mass scale, M.In this case the parameters comparing in Lp are not the directly measured parameters, but they can be correlated to the actual parameters by evaluating some observable quantity. From this point of view one could avoid the problems mentioned above by choosing a different point of renormalization. As far as the vacuum polarization is concerned, Πµν gives rise to the following correction of the photon propagator (illustrated in Fig. 2.7.4)

−igµν −igµλ −igρν D (q)= + ie2Πλρ(q) + ··· (2.184) µν q2 q2 q2

Fig. 2.7.4 - The loop expansion for the photon propagator. from which

−igµν −igµλ −igρν D (q)= + (ie2)(qλqρ − gλρq2)Π(q2) + ··· µν q2 q2 q2 −igµν qµqν = 1 − e2Π(q2) − i e2Π(q2)+··· (2.185) q2 q4

We see that the one loop propagator has a divergent part in gµν , and also divergent and ﬁnite pieces in the term proportional to the momenta. Therefore the propagator is not any more in the Feynman gauge. It follows that we should add to Lp

1 µν 1 µ 2 1 µ ν L = − Fµν F − (∂µA ) = A gµν A (2.186) 2 4 2 2

66 the two counterterms − C µν E µ 2 1 µν 1 µ 2 E C µ 2 (L ) = − Fµν F − (∂µA ) = −C Fµν F + (∂µA ) − (∂µA ) 2 ct 4 2 4 2 2 (2.187) As for the electron propagator, we can look at these two contributions as pertur- bations to the free lagrangian, and evaluate the corresponding Feynman rules, or evaluate the effect on the propagator, and then expand in the counter terms. The effect of these two terms to the equation which defines the propagator in momentum space is 2 νλ λ [(1 + C)q gµν − (C − E)qµqν]D (q)=−igµ (2.188) We solve this equation by putting

2 2 Dµν (q)=α(q )gµν + β(q )qµqν (2.189)

Substituting in the previous equation we determine α e β. The result is i i C − E α = − ,β= − (2.190) q2(1 + C) q4 (1 + C)(1 + E) The free propagator, including the corrections at the ﬁrst order in C and E is

−igµν qµqν Dµν (q)= (1 − C) − i (C − E) (2.191) q2 q4 and the total propagator

−igµν qµqν D (q)= [1 − e2Π(q2) − C] − i [e2Π(q2)+C − E] (2.192) µν q2 q4 We can now choose E = 0 and cancel the divergence by the choice e2 m C = − + F (2.193) 6π2 3 µ In fact we are free to choose the ﬁnite term of the gauge ﬁxing, since this choice does not change the physics. This is because the terms proportional to qµqν , as it follows from the gauge invariance, and the conservation of the electromagnetic current. For instance, if we have a vertex with a virtual photon (that is the vertex is connected to an internal photon line) and two external electrons, the term proportional to qµqν is saturated with u¯(p )γµu(p),q= p − p (2.194) The result is zero, bay taking into account the Dirac equation. Let us see what the full propagator says about the mass of the photon. We have

−igµν qµqν −igµν qµqν D (q)= 1 − F − e2Πf − i F + e2Πf ≡ [1 − Π]˜ − i Π˜ µν q2 3 q4 3 q2 q4 (2.195)

67 where 2 f Π=˜ e Π + F3 (2.196) At the second order in the electric charge we can write the propagator in the form −i qµqν Dµν (q)= gµν + Π(˜ q) (2.197) q2(1 + Π(˜ q)) q2 and we see that the propagator has a pole at q2 =0,sinceΠ(˜ q2) is ﬁnite for q2 → 0. Therefore the photon remains massless after renormalization. This is part of a rather general aspect of renormalization which says that if the regularization procedure respects the symmetries of the original lagrangian, the symmetries are preserved at any perturbative order. An exception is the case of the anomalous symmetries, which are symmetries at the classical level, but are broken by the quantum corrections. All the anomalous symmetries can be easily identiﬁed in a given theory. Also in this case we will require that at the physical pole, q2 → 0, the propagator has the free form, that is Π(0)˜ = 0 (2.198) from which 2 f F3 = −e Π (0) (2.199) For small momenta we can write dΠf (q) Π(˜ q) ≈ e2q2 (2.200) dq2 q2=0

f since F3 is a constant. Using the expression for Π from the previous Section, we get 1 γ 1 m2 1 1 q2 f q ≈− dz z2 − z 2 ··· Π ( ) 2 + log 2 + 2 (1 ) 2 + (2.201) 2π 6 6 2πµ 2π 0 m and e2q2 Π(˜ q) ≈ (2.202) 60π2m2 from which 2 2 gµν e q D = −i 1 − + gauge terms (2.203) µν q2 60π2m2 The ﬁrst term, 1/q2, gives rise to the Coulomb potential, e2/4πr. Therefore this is modiﬁed by a constant term in momentum space, or by a term proportional to the delta function in the real space d4q − − 1 e2 e2 e4 ∆ = e2 dt e iq(x1 x2) − = − − δ3(r) 12 (2π)4 q2 60π2m2 4πr 60π2m2 (2.204)

68 This modiﬁcation of the Coulomb potential modiﬁes the energy levels of the hydro- gen atom, and it is one of the contributions to the Lamb shift, which produces a splitting of the levels 2S1/2 and 2P1/2. The total Lamb shift is the sum of all the self-energy and vertex corrections, and turns out to be about 1057.9 MHz.The contribution we have just calculated is only −27.1 MHz, but it is important since the agreement between experiment and theory is of the order of 0.1 MHz. We have now to discuss the vertex corrections. We have seen that the divergent (1) contribution is Λµ (p ,p), and this is proportional to γµ. The counter term to add to interacting part of Lp Lint − ¯ µ p = eψγµψA (2.205) is ¯ µ (L3)ct = −eDψγµψA (2.206) The complete vertex is given by (see Fig. 2.7.5) 2 µ −ie γµ + e Λ + Dγµ (2.207)

(counter-term)

Fig. 2.7.5 - The one loop vertex corrections.

We ﬁx the counter term by e2 1 D = − + F (2.208) 8π2 2 with F2 the same as in eq. (2.7). The reason is that with this choice we get the equality of the wave function renormalization factors Z1 = Z2. This equality follows from the conservation of the current, and therefore we can use the arbitrariness in the ﬁnite part of the vertex counter term to guarantee it. The one-loop vertex is then µ 2 f F2 (Λ ) = γµ + e (Λ − γµ) (2.209) µ 8π2

One can see that this choice of F2 is such that for spinors on shell the vertex is the free one µ µ u¯(p)(Λ ) u(p)|pˆ=m =¯u(p)γ u(p) (2.210)

69 We will evaluate now the radiative corrections to the g − 2 of the electron. Here g is the gyromagnetic ratio, which the Dirac equation predicts to be equal to be two. To this end we ﬁrst need to prove the Gordon identity for the current of a Dirac particle µ µ p + p i ν u¯(p )γµu(p)=¯u(p ) + σµν q u(p) (2.211) 2m 2m con q = p − p. For the proof we start from

pγˆ µu(p)=(−mγµ +2pµ)u(p) (2.212) and γµpuˆ (p)=mγµ (2.213) Subtracting these two expressions we obtain pµ i ν γµu(p)= − σµν p u(p) (2.214) m m An analogous operation on the barred spinor leads to the result. We observe also that the Gordon identity shows immediately that g = 2, because it implies that the coupling with the electromagnetic ﬁeld is just

e µν σµν F (q) (2.215) 2m To evaluate the correction to this term from the one loop diagrams, it is enough to evaluate the matrix element

2 (2) e u¯(p )Λµ (p ,p)u(p) (2.216)

(1) in the limit p → p and for on-shell momenta. In fact Λµ contributes only to the terms in γµ, and in the previous limit they have to build up the free vertex, as implied by the renormalization condition. Therefore we will ignore all the terms proportional to γµ and we will take the ﬁrst order in the momentum q. For momenta (2) on shell, the denominator of Λµ is given by

[...]=m2(x + y) − m2x(1 − x) − m2y(1 − y)+2m2xy = m2(x + y)2 (2.217)

In order to evaluate the numerator, let us deﬁne

α µ Vµ =¯u(p )γ [ˆp (1 − y) − pxˆ + m] γ [ˆp(1 − x) − pˆ y + m] γαu(p) (2.218)

Usingpγ ˆ α = −γαpˆ+2pα, and an analogous equation forp ˆ ,wecanbringˆp to act on the spinor at the right of the expression, andp ˆ on the spinor at the left, obtaining α α α µ Vµ =¯u(p ) myγ +2(1− y)p − γ pxˆ γ [mxγα +2(1− x)pα − pˆ yγα] u(p) (2.219)

70 Making use of µ µ γαγ γα = −2γ (2.220) α γαγµγνγ =4gµν (2.221) µ α µ γαpγˆ pˆ γ = −2ˆp γ pˆ (2.222) we get 2 µ µ µ Vµ =¯u(p ) − 2m xyγ +2my(1 − x)(−mγ +2p ) − 2 µ − − µ µ − − 2 µ 4my p +2mx(1 y)( mγ +2p )+4(1 x)(1 y)m γ − 2y(1 − y)m2γµ − 4mx2pµ − 2m2x(1 − x)γµ − 2xym2γµ u(p) (2.223) and for the piece which does not contain γµ µ 2 µ 2 Vµ =4mu¯(p ) p (y − xy − x )+p (x − xy − y ) u(p) (2.224)

Therefore the relevant part of the vertex contribution is e2 1 1−x 4 (y − xy − x2) 2 (2) →− µ e u¯(p )Λµ (p ,p)u(p) 2 dx dy u¯(p ) p 2 16π 0 0 m (x + y) 2 µ (x − xy − y ) + p u(p) (2.225) (x + y)2 By changing variable z = x + y we have 1 1−x y − xy − x2 1 1 1 − x x dx dy = dx dz − (x + y)2 z z2 0 0 0 x 1 = [−(1 − x)logx + x − 1] 0 x2 x2 x2 1 1 = −x log x + x + log x − + − x = (2.226) 2 4 2 0 4 from which 2 2 (2) e e u¯(p )Λ (p ,p)u(p) →− u¯(p )[pµ + p ]u(p) (2.227) µ 16π2m µ Using the Gordon in this expression, and eliminating the further contribution in γµ we obtain 2 2 (2) ie ν e u¯(p )Λ (p ,p)u(p)| . . → u¯(p )σµν q u(p) (2.228) µ mom magn 16π2m Finally we have to add this correction to the vertex part taken at p = p,which coincides with the free vertex 2 2 ie ν pµ + pµ i e ν u¯(p )[γµ + σµν q ]u(p)=¯u(p )[ + 1+ σµν q ]u(p) (2.229) 16π2m 2m 2m 8π2

71 Therefore the correction is e e α → 1+ (2.230) 2m 2m 2π

Recalling that g is the ratio between S · B and e/2m,weget g α =1+ + O(α2) (2.231) 2 2π This correction was evaluated by Schwinger in 1948. Actually we know the ﬁrst three terms of the expansion 1 1 α α 2 α a = (g − 2) = − 0.32848 +1.49( )3 + ···= (1159652.4 ± 0.4) × 10−9 th 2 2 π π π (2.232) to be compared with the experimental value

−9 aexp = (1159652.4 ± 0.2) × 10 (2.233)

72 Chapter 3

Vacuum expectation values and the S matrix

3.1 In- and Out-states

For the future we will need a formulation of field theory showing that any S-matrix element can be reduced to the evaluation of the vacuum expectation value of a T - product of field operators. Such a formulation is called LSZ (from the authors l (Lehaman, Symanzik and Zimmermann). This formalism is based on a series of very general properties as displacement and Lorentz invariance. It is interesting to see how far one can go in determining the features of a field theory by using only invariance arguments. In the following we will work in the Heisenberg picture, that is the operators will be time-dependent and evolving with the full hamiltonian. Formally this representation can be obtained from eq. (1.217) by sending the free hamiltonian into the full hamiltonian. It should be also kept in mind that since no exact solution of an interacting field theory is known, we are not sure that the assumptions we are going to make are consistent. Said that, our assumptions are: 1. The eigenvalues of the four-momentum lie within the forward light cone

2 µ 0 p = p pµ ≥ 0,p≥ 0 (3.1)

2. There exists a nondegenerate Lorentz invariant states of minimum energy. This is called the vacuum state

Φ0 ≡|0 (3.2) By convention, we assume that the corresponding energy is zero

P0|0 =0 (3.3) From eq. (3.1) we get also

P |0 =0→ Pµ|0 =0 (3.4)

73 Since pµ = 0 is a Lorentz invariant condition, it follows that Φ0 appears as the vacuum states in all the Lorentz frames. 3. For each stable particle of mass m, there exists a stable stae of single particle

Φ1 ≡|p (3.5)

2 2 where Φ1 is a momentum eigenstate with eigenvalue pµ such that p = m . 4. Neglecting the complications from the zero mass states we can add a further requirement: the vacuum and the single particle states give a discrete spectrum in pµ. For instance, the case of the π mesons is given in Fig. 3.1. . E

xxx

xxx Continuum

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx 2m single particle state

xx xxm

xx Vacuum

Fig. 3.1 - Energy-momentum spectrum for particles of mass m.

Remember that the pion is not a stable particle and decays into µ +¯νµ with a lifetime of the order of ≈ 2 · 10−8 sec., but this number is very large with −24 respect to the natural decay frequency 1/mπ ≈ 5·10 sec. (mπ ≈ 140 MeV). Therefore, by neglecting the weak interactions (responsible for the decay) we can assume that the pion is stable. The same argument can be applied to quarks and leptons that will be the main actors in the following. However we mentioned also the pions, to say that the arguments presented here can be applied (in the appropriate range of energy) also to composite particles. Notice that this assumption is very much in the spirit of perturbation theory. That is we introduce a ﬁeld for each particle, assuming that the interactions do not modify the spectrum in a violent way. The main tools of investigation in the ﬁeld of elementary particle physics are the scattering experiments. Ideally these experiments are realized by preparing at

74 t = −∞ a system of free particles. That is assuming that they do not interact each other (this requires that the particles are well separated in space). These particles will interact with a target at some finite time. After that scattering process we imagine that at t =+∞ the emerging particles will behave again as free particles. In real life the times t = ±∞ correspond to laboratory time. That is we prepare the beam at some finite time −T , the scattering process will occur around the time t = 0 (assuming this as the time scale) and then we will measure the outgoing particles at a time T . The ideal description considered before can be considered as correct if the interaction time ∆t around t = 0 is much smaller that T . For instance, imagine an electron prepared in an eigenstate of momentum p, impinging over an atom. Here there is a further simplification that we are doing since in a finite time or a finite volume we cannot prepare a state of definite momentum. This is one of the reasons why analyzing a scattering process it is always convenient to quantize in a box of the dimension of the experiment itself (that is about the distance between the source of the beam and the detector). The same consideration holds for the energy, that is one has always to remember that the experiments is taking place between to finite times −T and T . Having prepared the electron in the initial state at the time −T , the particle will travel since it will reach the target where it will be subject to the interaction with the atoms of target itself. The interaction time will be of the order of the time necessary to cross the atomic dimensions for the case of a single scattering, or, at maximum, of the order of the thickness of the target for a multiple scattering. After that we will let again the electron to move freely since it will reach the detector where we will measure its properties as momentum and polarization. In conclusion we can think to describe the particles at t = ±∞ in terms of free fields, except that they will be subject to the self-interaction which, as we know, cannot be neglected. The free fields that we are going to use will be denoted by φin(x)for t →−∞and by φout(x)fort → +∞. The interacting field φ(x)) can be thought of as constructed in terms of these free fields operators and such that at t →±∞it reduces to φout(x)orφin(x). In order that the in- and out-fields describe correctly the incoming and outgoing free particles we have to require some properties. In particular we will require

1. φin(x) must transform as the corresponding φ(x) with respect to the symmetries of the theory. In particular, with respect to translation it must satisfy

∂φin(x) [Pµ, ]=−i (3.6) ∂xµ equivalent to require iP · a −iP · a φin(x + a)=e φin(x)e (3.7)

2. φin(x) must satisfy the free Klein-Gordon equation (for a scalar ﬁeld, or the Dirac free equation for a spinor ﬁeld, etc.) corresponding to the physical mass m 2 ( + m )φin(x) = 0 (3.8)

75 Notice that this requirement implies that we have taken into account the self- interactions of the ﬁeld which are responsible for the mass renormalization (we will mention the wave function renormalization later).

The same properties are required for the out-fields. By using these two properties it follows at once that the in-field creates from the vacuum the physical one particle state from the vacuum. In fact, let |p an eigenstate ofPµ with eigenvalue pµ.We have ∂ −i p|φ (x)|0 = p|[Pµ,φ (x)]|0 = pµ p|φ (x)|0 (3.9) ∂xµ in in in By applying −i∂/∂xµ once more, we find

2 − p|φin(x)|0 = p p|φin(x)|0 (3.10) and using the Klein-Gordon equation

2 2 (p − m ) p|φin(x)|0 = 0 (3.11)

We see that the only state that φin(x) can create out of the vacuum is the one with 2 2 pµ such that p = m . From our starting hypotheses we have that this is the state of single particle with mass m. Since the in- and out-fields are free fields we can apply all the corresponding formalism. In particular they can be Fourier expanded 3 † ∗ φin(x)= d k ain(k)fk(x)+ain(k)fk (x) (3.12) with 1 −ikx f (x)= e ,ω = k2 + m2 (3.13) k 3 k (2π) 2ωk Inverting this relation, we find 3 ∗ − ∗ ain(k)=i d x fk (x)∂0φin(x) ∂0fk (x)φin(x) (3.14) from which 3 ∗ ∂ ∂φin(x) ∂ ∗ ∂φin(x) [Pi,a (k)] = d x f (x) − f (x) in k ∂x0 ∂xi ∂x0 k ∂xi ∂f∗(x) ∂f∗(x) 3 k ∂ ∂ k = − d x φin(x) − φin(x) ∂xi ∂x0 ∂x0 ∂xi = −kiain(k) (3.15)

∗ Since fk (x)andφin(x) both satisfy the free Klein-Gordon equation, we see that the operators deﬁned in eq. (3.14) are time independent. This follows immediately, by

76 recalling that if φ1(x)andφ2(x) are both solutions of the Klein-Gordon equation, then the four-vector ∂φ2 ∂φ1 jµ = φ − φ (3.16) 1 ∂xµ ∂xµ 2 is a conserved current, and therefore the charge 3 Q = d xj0 (3.17) is a constant of motion. Using this, it follows ∂ ∂φ (x) ∂ ∂φ (x) P ,a k d3x f ∗ x in − f ∗ x in [ 0 in( )] = k ( ) 0 0 0 k ( ) 0 ∂x ∂x ∂x ∂x ∗ ∗ ∂f (x) ∂ ∂ ∂f (x) = − d3x k φ (x) − k φ (x) ∂x0 ∂x0 in ∂x0 ∂x0 in = −k0ain(k) (3.18)

In conclusion [Pµ,ain(k)] = −kµain(k) (3.19) Also, by hermitian conjugation we get † † [Pµ,ain(k)] = kµain(k) (3.20) † It follows that ain(k) creates states of four-momentum kµ whereas ain(k) annihilates states with momentum kµ. For instance,

† † | † † | † † | Pµain(k1)ain(k2) 0 =[Pµ,ain(k1)ain(k2)] 0 =(k1 + k2)µain(k1)ain(k2) 0 (3.21) † | † | − † | Pµain(k1)ain(k2) 0 =[Pµ,ain(k1)ain(k2)] 0 =(k2 k1)µain(k1)ain(k2) 0 (3.22) By using the condition on the spectrum of the momentum operator we get also ain(k)|0 = 0 (3.23) since, otherwise, this state would have a negative eigenvalue of the energy. We get also p1, ···,pM ;in|k1, ···,kN ;in = 0 (3.24) unless M = N and the set (p1, ···,pM ) is identical to the set (k1, ···,kN ). Let us now look for a relation between the in- and the φ-ﬁeld. The ﬁeld φ(x) will satisfy an equation of the type

( + m2)φ(x)=j(x) (3.25) where j(x) (the scalar current) contains the self-interactions of the ﬁelds and possible interactions with other ﬁelds in the theory. We need to solve this equation with the

77 boundary conditions at t = ±∞ relating φ(x) with the in- and out-fields. Notice that in writing the previous equation we have included the effects of mass renormalization inside the current j(x), but we know that at the same time we have also a wave function renormalization. Therefore asymptotically the field√ φ(x) cannot√ be the same as the in- and out-fields but it will differ by a factor Z,where Z is the wave function renormalization. Therefore, the required asymptotic conditions are √ √ lim φ(x) = lim Zφin(x), lim φ(x) = lim Zφout(x) (3.26) t→−∞ t→−∞ t→+∞ t→+∞

The solution for φ(x)isthen √ 4 2 φ(x)= Zφin(x)+ d y∆ret(x − y; m )j(y) (3.27)

2 We will see in a moment the meaning of Z in this context. The function ∆ret(x; m ) is the retarded Green’s function, deﬁned by

2 2 4 ( + m )∆ret(x; m )=δ (x) (3.28) such that 2 ∆ret(x; m )=0, for x0 < 0 (3.29) The need of Z in this context, is because we want the the in-field properly normalized in order to produce single particle states. In fact, both 1|φin(x)|0 e 1|φ(x)|0 have the same x-dependence (they must be both proportional to exp(ipx), see eq. (3.6)), but φ(x) may create out of the vacuum states other than the single particle state, and therefore the normalization of the two previous matrix elements cannot be the same. It can be shown that the asymptotic condition (3.26), cannot hold as an operator statement. This is because it is not possible to isolate the operator φ(x)fromthe scalar current j(x)att = −∞, since this includes also the self-interactions. However the statement can be given in a weak form for the matrix elements. The correct way of doing it is first to smear out the field over a space-like region f 3 ∗ ∗ φ (t)=i d x (f (x, t)∂0φ(x, t) − ∂0f (x, t)φ(x, t)) (3.30) with f(x, t) an arbitrary normalizable solution of the Klein-Gordon equation

( + m2)f(x) = 0 (3.31)

78 Similar considerations can be made for the out-field. In particular we can find a relation between the in- and out-fields √ 4 2 φ(x)= Zφout(x)+ d y∆adv(x − y; m )j(y) (3.33) where 2 ∆adv(x; m )=0, for x0 > 0 (3.34) 2 2 4 ( + m )∆adv(x; m )=δ (x) (3.35) is the advanced Green’s function. By comparing this expression with eq. (3.27) we find √ √ 4 2 Zφin(x)= Zφout(x)+ d y∆(x − y; m )j(y) (3.36) with 2 2 2 ∆(x − y; m )=∆adv(x − y; m ) − ∆ret(x − y; m ) (3.37) The construction of the out-operators of creation and annihilation goes along the same lines discussed for the in-operators.

3.2 The S matrix

We have shown in the previous Section that an initial state of n non-interacting particles can be described by

| ··· † ··· † | ≡| p1, ,pn;in = ain(p1) ain(pn) 0; in α;in (3.38) where α stands for the all collection of the quantum numbers of the initial state. After the scattering process we are interested to evaluate the probability amplitude for the ﬁnal state to be a state of non-interacting particles described by the out- operators | ··· † ··· † | ≡| p1, ,pn;out = aout(p1) aout(pn) 0; out β;out (3.39) The probability amplitude is given by the S matrix elements

Sβα = β;out|α;in (3.40)

We can also deﬁne an operator S transforming the in-states into the out-states

β;in|S = β;out| (3.41) such that Sβα = β;in|S|α;in (3.42) From the deﬁnition we can derive some important properties for the operator S

|0; out = |0; in ≡|0 (3.44)

2. In an analogous way the stability of the one particle state requires

|p;in = |p;out (3.45)

3. The operator S maps the in- to the out-ﬁelds

−1 φin(x)=Sφout(x)S (3.46)

In fact, let us consider the following matrix element

β;out|φout|α;in = β;in|Sφout|α;in (3.47)

But β;out|φout is an out-state and therefore

β;out|φout = β;in|φinS (3.48)

Then

which shows the relation (3.46).

4. The S is unitary. This follows from the very deﬁnition (3.41) implying

S†|α;in = |α;out (3.50)

from which † β;in|SS |α;in = β;out|α;out = δαβ (3.51) showing that SS† = 1 (3.52) From eq. (3.50) and the unitarity of S we get

−1 |α;in = S† |α;out = S|α;out (3.53)

5. S is Lorentz invariant, as it follows from the fact that transform in- into out- ﬁelds having the same Lorentz properties. For the same reason the S matrix is invariant under any symmetry of the theory.

80 3.3 The reduction formalism

In this Section we will describe a systematic way of evaluating the S matrix elements in terms of the vacuum expectation values of T -products of ﬁeld operators. Let us start considering the following S matrix element

Sβ,αp = β;out|α, p;in (3.54) where |α, p;in is the state of a set of particles α plus an additional incoming particle of momentum pµ. Here we will consider the case of scalar particles, but the formalism can be easily extended to other cases. We want to show that, by using the asymptotic condition, it is possible to extract the particle of momentum pµ from the state by introducing the corresponding ﬁeld operator. Let us consider the following identities | | † | β;outα, p;in = β;outain(p) α;in | † | | † − † | = β;outaout(p) α;in + β;outain(p) aout(p) α;in = β − p;out|α;in 3 ∂ − i d x fp(x) β;out|φ (x) − φ (x)|α;in ∂x0 in out ∂ − fp(x) β;out|φ (x) − φ (x)|α;in (3.55) ∂x0 in out where we have used the adjoint of eq. (3.14). We have denoted |β − p;out the out- state obtained by removing the particle with momentum pµ, if present in the state, otherwise it must be taken as the null vector. As we know the integral appearing in the previous expression is time-independent since fp(x), φin(x)andφout(x)are solutions of the Klein-Gordon equation. Therefore we are allowed to make the following substitutions (in the weak sense, since we are taking matrix elements) 1 lim φin(x) = lim √ φ(x) x0→−∞ x0→−∞ Z 1 lim φout(x) = lim √ φ(x) (3.56) x0→+∞ x0→+∞ Z obtaining β;out|α, p;in = β − p;out|α;in i 3 ∂ + √ lim − lim d x fp(x) β;out|φ(x)|α;in Z x0→+∞ x0→−∞ ∂x0 ∂ − fp(x) β;out|φ(x)|α;in (3.57) ∂x0 By using the identity +∞ ∂ ∂ ∂ 4 − d x 0 g1(x) 0 g2(x) 0 g1(x) g2(x) = −∞ ∂x ∂x ∂x

81 3 ∂ ∂ = lim − lim d x g1(x) g2(x) − g1(x) g2(x) x0→+∞ x0→−∞ ∂x0 ∂x0 +∞ 4 = d x [g1(x)¨g2(x) − g¨1(x)g2(x)] (3.58) −∞ we can write

β;out|α, p;in = β − p;out|α;in i 4 + √ d x β;out| fp(x)φ¨(x) − f¨p(x)φ(x) |α;in (3.59) Z

By using the Klein-Gordon equation for fp(x)weget

| − | β;out α, p;in = β p;outα;in i 4 2 2 + √ d x β;out| fp(x)φ¨(x) − (∇ − m )fp(x) φ(x) |α;in (3.60) Z Integrating by parts

| − | β;outα, p;in = β p;outα;in i 4 2 + √ d xfp(x)( + m ) β;out|φ(x)|α;in (3.61) Z We can iterate this procedure in order to extract all the in- and out-particles from the states. Let us see also the case in which we want to extract from the previous matrix element an out-particle. This exercise is interesting because it shows how the T -product comes about. Let us suppose that the out state is of the type β =(γ,p), with p the momentum of a single particle. We want to extract this particle from the state

Using again the time independence of the integral we have

| − | γ;out [φout(y)φ(x) φ(x)φin(y)] α;in 1 = √ lim − lim γ;out|T (φ(y)φ(x))|α;in (3.63) Z y0→+∞ y0→−∞

82 and eq. (3.58)

More generally if we want to remove all the particles in (q1, ···,qm)andin(p1, ···,pn) with the condition pi = qj we get

p , ···,pm;out|q , ···,qn;in 1 1 m+n m,n √i 4 4 ∗ 2 2 · = d xid yjfqi (xi)fpj (yj)( + m )xi ( + m )yj Z i,j=1

0|T (φ(y1) ···φ(yn)φ(x1) ···φ(xm))|0 (3.66)

If some of the qi’s are equal to some of the pj’s, we can reduce further the matrix elements of the type β;out|α−p;in . Therefore we have shown that it is possible to evaluate all the S matrix elements once one knows the vacuum expectation values of the T -products (called also the n-point Green’s function). The result obtained here can be easily extended to generic fields simply by substituting to the wave functions fp(x) the corresponding solutions of the wave equation (for instance, for spinor fields, spinors times plane waves) and to the Klein-Gordon operator the corresponding wave / operators (for the spinor√ case the√ Dirac operator (i∂ − m)).√ In the Dirac case one halso to change i/ Z to −i/ Z2 for fermions and to +i/ Z2 for antifermions, where Z2 is the wave function renormalization for the spinor field.

83 We will show in the following that a simple technique to evaluate perturbatively the vacuum expectation value of n ﬁelds, a n-point Green’s function, can be obtained in the context of the path-integral formulation of quantum mechanics.

84 Chapter 4

Path integral formulation of quantum mechanics

4.1 Feynman’s formulation of quantum mechanics

In 1948 Feynman gave a formulation of quantum mechanics quite different from the standard one. This formulation did not make use of operators and of Schrödinger equation, but rather it was giving an explicit expression for the quantum-mechanical amplitude, although using a mathematical formalism which was far from being well defined, which was obtained directly from the physical idea that the probability amplitude to go from one state to another can be obtained by summing together all the possible ways with an appropriate weight. This was as apply the ordinary composition laws in probability theory, to the probability amplitudes (remember that the probability is the square of the amplitude). The key observation to get the appropriate weight was taken by the Dirac’s book where it was shown that, for one degree of freedom, the probability amplitude

q,t+∆t|q, t (4.1) for ∆t infinitesimal, is given by the eiS where S is the classical action to go from q at the time t to q at the tome t +∆t. In fact we will show that the Feynman’s formulation can be derived from the usual formulation of quantum mechanics. Although the path integral formulation (as the Feynman’s approach is usually called) is just a different way of formulating quantum mechanics, at the beginning of the 70’s it turned out to be a crucial tool for the quantization of gauge theories. Also it is one the few methods which can be used for studying situations outside the perturbative approach. In order to derive the Feynman’s formulation we will start studying a quantum system described by one degree of freedom, and we will see later how this can be extended also to infinite degrees of freedom as needed in quantum field theory. First of all we notice that the quantum mechanical problem is to evaluate

85 the time evolution, which is given by the propagator. This can be evaluated in any basis, for instance in conﬁguration space, its matrix elements are given by

q,t|q, t ,t ≥ t (4.2)

It will be convenient to work in the Heisenberg representation, that is using time- dependent operators q(t). The states of the previous equation are deﬁned as eigenstates of the operator q(t)atthetimet,thatis

q(t)|q, t = |q, t q (4.3)

It follows immediately that the relation between these states at diﬀerent times is given by |q, t = eiHt|q, 0 (4.4) for a time independent hamiltonian. The matrix element (4.2) can be evaluated by slicing the time interval t − t in inﬁnitesimal pieces and using the completeness relation. we get q ,t|q, t = lim dq ···dqn− q ,t|qn− ,tn− qn− ,tn− |qn− ,tn− · n→∞ 1 1 1 1 1 1 2 2

··· q1,t1|q, t (4.5) where tk = t + k, t0 ≡ t, tn ≡ t (4.6) Therefore n−1 n−1 q ,t|q, t = lim dqk qk ,tk |qk,tk (4.7) n→∞ +1 +1 k=1 k=0 We have still to evaluate the matrix element (4.2), but now for an inﬁnitesimal time interval . let us use again the completeness in momentum space with states taken at a time t˜ in between tk e tk+1:

qk+1,tk + |qk,tk = dp qk+1,tk + |p, t˜ p, t˜|qk,tk (4.8)

The mixed matrix elements can be written as

−iH(p, q)(tk + − t˜) qk+1,tk + |p, t˜ = qk+1, 0|e |p, 0 (4.9) (remember /h = 1). making use of the canonical commutation relations for p and q, the hamiltonian can be written in such a way that all the operators q are at the left of all the operators p. The result will be denoted by the symbol H+(q, p). By doingsoweget qk+1,tk + |p, t˜ qk+1, 0|p, 0 1 − iH+(qk+1,p)(tk + − t˜)

−iH+(qk+1,p)(tk + − t˜) qk+1, 0|p, 0 e (4.10)

86 The function H+(q, p) is obtained by substituting in H+(q, p) the operators q and p with their eigenvalues. Notice that although H(q, p)andH+(q, p) are the same operator, H(q, p)eH+(q, p) are, in general, diﬀerent functions. In this way we get 1 i(pqk+1 − H+(qk+1,p)(tk+1 − t˜)) qk+1,tk + |p, t˜ √ e (4.11) 2π and 1 i(−pqk − H−(qk,p)(t˜− tk)) p, t˜|qk,tk √ e (4.12) 2π where H− is the eigenvalue of the operator H−(p, q) deﬁned by putting the operators q to the right of p by using the canonical commutation relations. By choosing t˜=(tk + tk+1)/2 and the previous two equations, we get dp i(p(qk − qk)/ − Hc(qk ,qk,p)) qk ,tk + |qk,t e +1 +1 (4.13) +1 2π where 1 Hc(qk ,qk,p)= (H (qk ,p)+H−(qk,p)) (4.14) +1 2 + +1 Since in the limit → 0wehave

qk+1,tk + |qk,tk →δ(qk+1 − qk) (4.15) it is reasonable to deﬁne a variable velocity as

qk+1 − qk q˙k = (4.16) and write Hc(qk+1,qk,p) ≡ H(qk,p) (4.17)

In the same limit H(qk,p) is nothing but the hamiltonian of the system evaluated in the phase space. Therefore dp iL(qk,p) qk ,tk + |qk,tk e (4.18) +1 2π where L(qk,p)=pq˙k − H(qk,p) (4.19) is the lagrangian in phase space. Coming back to (4.7) and using (4.18) we obtain

n−1 n−1 n−1 i L(qk,pk) dpk k q ,t|q, t = lim dqk e =0 (4.20) n→∞ 2π k=1 k=0

87 In the limit, the argument of the exponential gives the integral between t and t of the lagrangian in the phase space, therefore we get the canonical action

t S = dt L(q, p) (4.21) t

In the limit n →∞, the multiple integral in eq. (4.20) is called the functional integral and we will denote it symbolically as

t q,t i dtL(q, p) q,t|q, t = dµ(q(t))dµ(p(t))e t (4.22) q,t where dp(t) dµ(q(t)) = dq(t),dµ(p(t)) = (4.23) t

4.2 Path integral in conﬁguration space

The path integral in eq. (4.20) can be easily written as a functional integral in conﬁguration space when the hamiltonian of the system is of the type

p2 H(q, p)= + V (q) (4.24) 2m In such a case p2 H (q, p)=H−(q, p)=H(q, p)= + V (q) (4.25) + 2m The momentum integration can be done naively n−1 q ,t|q, t = lim dqk n→∞ k=1 n−1 dpk − − 2 − · ei[pk(qk+1 qk) pk/2m (V (qk+1)+V (qk))/2] = 2π k=0 n n−1 n−1 2 1 2mπ (−2m/i)(qk − qk) /4 = lim dqk e +1 n→∞ 2π i k=1 k=0 − · e i(V (qk+1)+V (qk))/2 =

88 n n− m 1 = lim dqk n→∞ 2iπ k=1 n−1 − 2 − · ei[m((qk+1 qk)/) /2 (V (qk+1)+V (qk))/2] (4.26) k=0 with the result n n−1 m q ,t|q, t = lim dqk n→∞ 2iπ k=1 n−1 2 i [m((qk+1 − qk)/) /2 − (V (qk+1)+V (qk))/2] · e k=0 (4.27)

We have also n−1 2 t m qk − qk V (qk )+V (qk) lim +1 − +1 = dtL(q, q˙) (4.28) n→∞ 2 2 t k=0 where L(q, q˙) is the lagrangian in conﬁguration space. Therefore we will write

t q,t i dtL(q, q˙) q,t|q, t = D(q(t))e t (4.29) q,t with the functional measure given by

n n− m 1 D(q(t)) = lim dqk (4.30) n→∞ 2iπ k=1

It is the expression (4.29) (really deﬁned by (4.27)) which was originally formulated by Feynman. In fact it has a simple physical. But before discussing the interpretation let us notice that for N degrees of freedom and with hamiltonians of the type N 2 pi H(qi, pi)= + V (qi) i =1,...,N (4.31) 2m i=1 the expression (4.29) can be easily generalized. However for arbitrary system one has to start with the formulation in the phase space of the previous Section, which also can be easily generalized to an arbitrary number of degrees of freedom. For discussing the interpretation of the expression (4.29), let us recall the boundary conditions q(t)=q, q(t)=q (4.32)

89 .

q q' Fig. 4.2.1 - The Feynman deﬁnition for the path integral.

The expression in the exponent of (4.29) is nothing but the action evaluated for a continuous path starting from q at time t and ending at q at time t.Thispathis approximated in eq. (4.27) by many inﬁnitesimal straight paths as shown in Fig. 4.2.1. Therefore the interpretation is that in order to evaluate the amplitude to go from (q, t)to(q,t) is obtained by multiplying together the amplitudes corresponding to the inﬁnitesimal straight paths. Each of these amplitudes (neglecting the normalization factor m/2iπ) is given by the exponential of i times the action evaluated along the path. Furthermore we need to sum (or integrate functionally) over all the possible paths joining the two given end points. In a more concise way one says that the amplitude to from one point to another is obtained by summing over all the possible paths joining the two points with a weight proportional to the exponential of iS,whereS is the classical action evaluated along the path.

90 4.3 The physical interpretation of the path integral

Feynman arrived to the path integral formulation in configuration space on a more physical basis. First of all he knew that the amplitude for infinitesimal time differ- ences is proportional to the exponential of iS,withS the classical action evaluated along the infinitesimal path. From that one can derive directly the final formula in the following way. Let us consider the classical double slit experiment, as illustrated in Fig. 4.3.1, where S is the electron source. The electrons come out of the slit A and arrive to the screen C, where they are detected by D, going through one of the two silts A or B. By counting the electrons with the detectpr D we can determine, as a function of x, how many electrons arrive on C, and therefore the probability distribution of the electrons on the screen C. Under the conditions of Fig. 4.3.1, that is with both slits 1 and 2 open, one gets the distribution given in Fig. 4.3.2. In order to explain this distribution it would seem reasonable to make the following assumptions: . D

1 x

AB C Fig. 4.3.1 - The double slit experiment.

. P

x Fig. 4.3.2 - The experimental probability distribution, P , with both slits open.

• Each electron must go through the slit 1 or through the slit 2.

91 • As a consequence the probability for the electron to reach the screen at the point x must be given by the sum of the probability to reach x through 1 with the probability to go through 2. These hypotheses can be easily checked by closing one of the two slits and doing again the experiment. The results obtained in the two cases are given in Fig. 4.3.3. . P 2 P1

Fig. 4.3.3 - The experimental probability distributions, P1 and P2 obtained keeping open the sli1 1 or the slit 2 respectively.

We see that P = P1 + P2, therefore one of our hypotheses must be wrong. In fact, quantum mechanics tells us that we can say that an electron went through one of the two slits with absolute certainty, only if we detect its passage through the slit. Only in this case one would get P = P1 + P2. In fact, this is the result that one obtains when doing the experiment with a further detector telling us which of the holes have been crossed by the electron. This means that with this experimental setting the interference gets destroyed. On the contrary the experiment tells us that when we do not detect the hole crossed by the electron, the total amplitude for the process is given by the sum of the amplitudes corresponding to the two different possibilities, that is | |2 | |2 | |2 | |2 | |2 | |2 P = A = A1+A2 = A1 + A2 +A1A2+A1A2 = A1 + A2 = P1+P2 (4.33) The two expressions differ for a term which is not positive definite and which is the responsible for the interference pattern. Therefore, quantum mechanics tells us that the usual rules for combining together probabilities are correct, but the result that we get follows from our ignorance about the path followed by the particle, unless we measure it. That is it does not make sense, under such circumstances, define a concept as the path followed by a particle. The Feynnman’s interpretation is a little bit different although at the end the theories turn out to be equivalent. He says that it is still possible to speak about trajectories of the electron if one gives up to the usual rules of combining probabilities. This point of view requires taking into account different type of alternatives according to different experimental settings. In particular one defines the following alternatives

92 1. Exclusive alternative - We will speak about exclusive alternatives when we use an apparatus to determine the alternative chosen by the system. For instance one we use a detector to determine the slit crossed by the electrons. In this case the total probability is given by the sum of the probabilities corresponding to the diﬀerent alternatives, P = Pi. (4.34) i

2. Interfering alternatives - This is the situation where the alternative chosen by the system is not detected by an experimental apparatus. In this case we associate to each alternative an amplitude Ai and assume that the total amplitude is given by the sum of all the amplitudes Ai A = Ai (4.35) i

The total probability for the process is evaluated as 2 2 P = |A| = | Ai| (4.36) i

Therefore the probability rules are strictly related to the experimental setting. There are situations in which one may have both type of alternatives. For instance, if we want to know the total probability of having an electron in C within the interval (a, b), having a detector in C and not looking at the slit crossed by the electrons. In this case the total probability will be given by

b 2 P = dx|A1(x)+A2(x)| (4.37) a Since the alternatives of getting the electron at diﬀerent points in C are exclusive ones. From this point of view it is well conceivable to speak about trajectories of the electron, but we have to associate to each possible path an amplitude and then sum all over the possible paths. For instance, suppose that we want to evaluate the amplitude to go from S to the screen C when the screen B is eliminated. This can be done by increasing the number of holes in the screen B. Then, if we denote by A(x, y) the amplitude for the electron to reach x going through the hole placed at adistancey from the center of B, the total amplitude will be given by A(x)= dyA(x, y) (4.38)

We can continue by using more and more screen of type B with more and more holes. That is we can construct a lattice of points between A and C and construct

93 all the possible paths from A to the point x on C joining the sites of the lattice, as shown in Fig. 4.3.4. By denoting with C the generic path we get that the total amplitude is given by A(x)= A(x, C) (4.39) C

. D

C 1 x

A C Fig. 4.3.4 - The lattice for the evaluation of equation (4.39).

From this point it is easy to get the path integral expression for the total amplitude, since each path is obtained by joining together many infinitesimal paths, and for each of these paths the amplitude is proportional to exp(iS)whereS is the action for the infinitesimal path. Therefore, for the single path C the amplitude will be given by exp(iS(C). This is the final result, except for a proportionality factor. This is easy to get if one starts from the phase space formulation. This is important because in many cases the simple result outlined above is not the correct one. For instance, consider the following lagrangian m L = q˙2f(q) − V (q) (4.40) 2 where f(q) is an arbitrary function of q. The canonical momentum is given by

p = mqf˙ (q) (4.41) and the corresponding hamiltonian is 1 H = p2 + V (q) (4.42) 2mf(q)

94 Notice that for qk ≈ qk+1 we have f(¯qk) ≈ (f(qk)+f(qk+1))/2, whereq ¯k =(qk + qk+1)/2. Therefore, form eq. (4.20) we get n−1 q ,t|q, t = lim dqk n→∞ k=1 n−1 dpk − − 2 − · ei[pk(qk+1 qk) pk/(2mf(¯qk)) (V (qk+1)+V (qk))/2] (4.43) 2π k=0 This expression is formally the same as the one in eq. (4.26) after sending m → mf(¯qk). Therefore we get n n−1 mf(¯qk) q ,t|q, t = lim dqk n→∞ 2iπ k=1 n−1 2 i [mf(¯qk)((qk+1 − qk)/) /2 − (V (qk+1)+V (qk))/2] · e k=0 (4.44)

We can rewrite this expression in a way similar to the one used in eq. (4.29)

t q,t i dtL(q, q˙) q,t|q, t = D(q(t))e t (4.45) q,t with L(q, q˙) given in (4.40), but the functional measure is now n n−1 mf(¯qk) D(q(t)) = lim dqk (4.46) n→∞ 2iπ k=1 The formal expression for the path integral is always the same, but the integration measure must be determined by using the original phase space formulation. In the continuum limit we have n− 1 1 n−1 log f(¯qk) 2 k lim f(¯qk) = lim e =0 = n→∞ n→∞ k=0 n− 1 1 t log f(¯qk) 1 2 δ(0) dt log f(q) = lim e k=0 = e2 t (4.47) n→∞ wherewehaveused n−1 t δhk lim = dt , lim = δ(th − tk) (4.48) n→∞ t n→∞ k=0

95 Therefore the result can be also interpreted by saying that the classical lagrangian gets modified by quantum effects as follows i L → L − δ(0) log f(q) (4.49) 2 The reason why we say that this is a quantum modification is because the second term must be proportional toh /, as it follows from dimensional arguments. An interesting question is how one gets the classical limit from this formulation. Let us recall that the amplitude for a single path is exp(iS(C)//h)(wherewehave put back the factorh /). To find the total amplitude we must sum over all the possible paths C. Since the amplitudes are phase factors the will cancel except for the paths which are very close to a stationary points of S. Around this point the phases will add coherently, whereas they will cancel going away from this point due to the fast variation of the phase. Roughly we can expect that the coherence effect will take place for those paths such that their phase differs from the stationary value, S(Cclass)//h, of the order of one. For a macroscopical particle (m ≈ 1grandtimesof about 1 sec) the typical action is about 1027/h, and the amplitude gets contribution only from the classical path. For instance, take a free particle starting from the origin at t = 0 and arriving at x0 at the time t0. The classical path (the one making S stationary) is t xclass(t)=x0 (4.50) t0 The corresponding value for S is t0 2 1 2 1 x0 S(xclass)= m q˙ = m (4.51) 2 0 2 t0

Let us now consider another path joining the origin with x0 as, for instance, t2 x(t)=x0 2 (4.52) t0 The corresponding action is 2 x2 S(x)= m 0 (4.53) 3 t0 and we get 2 1 x0 ∆S = S(x) − S(xclass)= m (4.54) 6 t0

Let us take t0 =1secandx0 = 1 cm. We get ∆S = m/6. For a macroscopic particle (say m = 1 gr) and recalling thath / ≈ 10−27 erg·sec, we get ∆S =1.6·1026/h and this non-classical path can be safely ignored. However, for an electron m ≈ 10−27 gr, and ∆S =/h/6. It follows that in the microscopic world, many paths may contribute to the amplitude.

96 4.4 The free particle

We shall now consider the case of the one-dimensional free motion. In this case we can easily make the integrations going back to the original deﬁnition (eq. (4.27). The expression we wish to evaluate is n−1 − 2 n n−1 i m((qk+1 qk)/) /2 m k q ,t|q, t = lim dqk e =0 (4.55) n→∞ 2iπ k=1 since the lagrangian is 1 L = mq˙2 (4.56) 2 The integrations can be done immediately starting from the gaussian integral ∞ 2 π − 2 dxeax + bx = − e b /4a, Re a ≤ 0 (4.57) −∞ a which allows us to evaluate the expression − 2 − 2 π − 2 dxea(x1 x) + b(x2 x) = − eab(x1 x2) /(a + b) (4.58) a + b let us start by integrating q1 − 2 − 2 iπ − 2 dq eim[(q2 q1) +(q1 q0) ]/(2) = eim(q2 q0) /(2(2) 1 m 1 2πi m 2 − 2 = eim(q2 q0) /(2(2) (4.59) m 2πi(2)

Multiplying by the exponential depending on q2 and integrating over this variable we get 1 2πi m 2 − 2 − 2 dq eim[(q3 q2) +(q2 q0) /2]/(2) m 2πi(2) 2 1 1 2πi m 2 2πi(2) 2 − 2 = eim(q3 q0) /(2(3)) m 2πi(2) 3m 3 1 2πi 2 m 2 − 2 = eim(q3 q0) /(2(3)) (4.60) m 2πi(3) By repeating this procedure, after n − 1 steps we ﬁnd n−1 − 2 n−1 i m((qk+1 qk)/) /2 k dqk e =0 k=1 n 1 2πi 2 m 2 − 2 = eim(qn q0) /(2(n)) (4.61) m 2πi(n)

97 The ﬁrst factor in the right hand side of this equation cancels out an analogous factor in the integration measure (see eq. (4.55))

1 m 2 − 2 − q,t|q, t = eim(q q) /(2(t t)) (4.62) 2πi(t − t) This expression can be generalized to a free system with N degrees of freedom N 2 N − i − 2 im (qi q ) /(2(t t)) m i q ,t|qi,t = e =1 ,i=1, ···,N (4.63) i 2πi(t − t) Notice also that lim q,t|q, t = δ(q − q) (4.64) t→t The classical path connecting (q, t)to(q,t) for the free particle is given by τ − t q (τ)=q + (q − q) (4.65) cl t − t The corresponding classical action is

t m m (q − q)2 t m (q − q)2 S q2 dτ dτ cl = ˙cl = 2 = (4.66) t 2 2 (t − t) t 2 (t − t) Therefore our result can be written as

1 m 2 q,t|q, t = eiScl (4.67) 2πi(t − t) We shall see that for all the lagrangians quadratic in the coordinates an in the momenta this result, that is that the amplitude is proportional to the exponential of iS with S the action evaluated along the classical path, is generally true. Of course, one could formulate quantum mechanics directly in terms of the path integral. It is not difficult to show that it is possible to recover all the results of the ordinary formulation and prove that one has complete equivalence. However we will not do it here, but we will limit ourselves to notice that the formulations agree in the free case. we take a fixed initial point of propagation, for instance (q, t)=(0, 0). Then we define the wave function as q,t ψ(q,t)= q,t|0, 0 = D(q(t))eiS (4.68) 0,0 For the free particle we get

2 1 mq m 2 i ψ(q, t)= e 2t (4.69) 2πit

98 This is indeed the wave function for a free particle in conﬁguration space, as it can be checked by going in the momentum space

2 2 1 1 mq 1 p 2 − m 2 − i m 2 2πit i t − ψ(p, t)= dq e ipqe 2t = e 2m = e iEt (4.70) 2πit 2πit m where E = p2/2m.

4.5 The case of a quadratic action

In this Section we will consider the case of a quadratic lagrangian for a single degree of freedom (it can be easily extended to many degrees of freedom) 1 1 L = mq˙2 + b(τ)˙qq − c(τ)q2 + d(τ)˙q − e(τ)q − f(τ) (4.71) 2 2 To evaluate the integral (4.29) let us start by the Euler-Lagrange equations associated to eq. (4.71) d [mq˙ + b(τ)q + d(τ)] − [b(τ)˙q − c(τ)q − e(τ)] = 0 (4.72) dτ that is mq¨ + bq˙ + d˙ + cq + e = 0 (4.73)

Let us denote by qcl(τ) the solution of the classical equations of motion satisfying the boundary conditions

qcl(t)=q, qcl(t )=q (4.74) We then perform the change of variables

η(τ)=q(τ) − qcl(τ) (4.75) which in terms of our original deﬁnition (4.29) means

ηk = qk − qclk,k=1, ···,n− 1 (4.76) Since this is a translation the functional measure remains invariant. The new variable η(τ) satisﬁes the boundary condition

η(t)=η(t) = 0 (4.77)

The lagrangian in terms of the new variables is given by 1 1 L(q)=L(q + η)=L(q )+ mη˙2 + bηη˙ − cη2 cl cl 2 2 + dη˙ − eη + mq˙clη˙ + bq˙clη + bηq˙ cl − cqclη (4.78)

99 Inside the action we can integrate by parts the terms containing both qcl and η.By using the boundary conditions for η we get 1 1 L(q)=L(q )+ mη˙2 + bηη˙ − cη2 + dη˙ − eη − mq¨ + bq˙ + cq η (4.79) cl 2 2 cl cl cl

Recalling the equations of motion for qcl and performing a further integration by parts we find 1 1 L(q)=L(q )+ mη˙2 + bηη˙ − cη2 + dη˙ − eη − −d˙ − e η cl 2 2 1 1 = L(q )+ mη˙2 + bηη˙ − cη2 (4.80) cl 2 2 Therefore we can write the path integral as follows t 1 2 1 2 0,t i ( mη˙ + bηη˙ − cη )dτ q,t|q, t = eiScl(q, q ) D(η(τ))e t 2 2 (4.81) 0,t Notice that here all the dependence on the variables q and q is in the classical action. In fact the path integral which remains to be evaluated depends only on t and t due to the boundary conditions satisfied by η. For the future purposes of extension of this approach to field theory, this term turns out to be irrelevant. However it is important for systems with a finite number of degrees of freedom and therefore we will give here an idea how one can evaluate it in general. First, let us notice that the term bηη˙ is a trivial one, since we can absorb it into a re-definition of c(τ), through an integration by part d 1 1 bηη˙ = bη2 − bη˙ 2 (4.82) dτ 2 2 and defining c˜ = c + b˙ (4.83) In this way we get t n−1 1 2 1 2 m 2 1 2 i mη˙ − cη˜ dτ = lim i (ηk − ηk) − c˜kηk n→∞ +1 t 2 2 k 2 2 =0 n−1 im 2 i 2 = lim 2ηk − 2ηkηk − c˜kηk n→∞ 2 +1 2 k=1 = lim iηT ση (4.84) n→∞ where we have used the boundary conditions on η (η0 = ηn = 0) and we have introduced the vector ⎡ ⎤ η1 ⎢ ⎥ ⎢ η2 ⎥ ⎢ ⎥ η = ⎢ · ⎥ (4.85) ⎣ · ⎦

ηn−1

100 and the matrix⎡ ⎤ ⎡ ⎤ 2 −10··· 000 c˜1 0 ··· 00 ⎢ ⎥ ⎢ ⎥ ⎢ −12−1 ··· 000⎥ ⎢ 0˜c2 ··· 00⎥ ⎢ ⎥ ⎢ ⎥ m ⎢ 0 −12··· 000⎥ ⎢ · · ··· · · ⎥ σ = ⎢ ⎥ − ⎢ ⎥ 2 ⎢ ·········⎥ 2 ⎢ · · ··· · · ⎥ ⎣ ⎦ ⎣ ⎦ 000··· −12−1 00··· cñ−2 0 000··· 0 −12 00··· 0˜cn−1 (4.86) Therefore t 1 1 0,t i ( mη˙2 + bηη˙ − cη2)dτ G(t,t) ≡ D(η(τ))e t 2 2 0,t n− n 1 T m 2 iη ση = lim dηk e (4.87) n→∞ 2πi k=1 Since σ is a symmetric matrix, it can be diagonalized by an orthogonal transformation T σ = U σDU (4.88) The eigenvectors of σ, ζ = Uη, are real and the Jacobian of the transformation is one, since det|U| =1.Weget n−1 n−1 n−1 n−1 T T 2 iη ση iζ σDζ π π dηk e = dζk e = = (−iσk) − k=1 k=1 k=1 det( iσ) (4.89) We get 1 1 m n πn−1 2 m 1 1 2 G(t,t) = lim = lim (4.90) n→∞ − n→∞ 2i n−1 − 2πi det( iσ) 2πi ( m ) det( iσ) Let us define n− 2i 1 f(t,t) = lim det(−iσ) (4.91) n→∞ m then m G(t,t)= (4.92) 2πif(t,t) In order to evaluate f(t,t), let us define (for j =1, ···,n− 1) ⎡ ⎤ ⎡ ⎤ 2 −10··· 000 c˜1 00··· 00 ⎢ ⎥ ⎢ ⎥ ⎢ −12−1 ··· 000⎥ ⎢ 0˜c2 0 ··· 00⎥ ⎢ ⎥ 2 ⎢ ⎥ ⎢ 0 −12··· 000⎥ ⎢ 00˜c3 ··· 00⎥ Pj = ⎢ ⎥ − ⎢ ⎥ ⎢ ·········⎥ m ⎢ ······· ·⎥ ⎣ ⎦ ⎣ ⎦ 000··· −12−1 000··· c˜j−1 0 000··· 0 −12 000··· 0˜cj (4.93)

101 and pj = det|Pj| (4.94) Clearly n− 2i 1 det(−iσ)=pn− (4.95) m 1

For the ﬁrst values of j, pj is given by

2 p1 =2− c˜1 m 2 p2 = 2 − c˜2 p1 − 1 m 2 p = 2 − c˜ p − p (4.96) 3 m 3 2 1 Therefore we ﬁnd the recurrency relation 2 pj = 2 − c˜j pj − pj− ,j=1, ···,n− 1 (4.97) +1 m +1 1 wherewedeﬁnep0 = 1. It is convenient to put this expression in the form

pj − 2pj + pj− c˜j pj +1 1 = − +1 (4.98) 2 m It is also convenient to introduce the ﬁnite diﬀerence operator

pj+1 − pj ∆pj = (4.99) atimeτ = t + j, and a function φ(τ) such that

lim pj = φ(τ) (4.100) →0 in the limit in which we keep τ ﬁxed. It follows dφ(τ) lim ∆pj = (4.101) →0 dτ The equation (4.98) can be written as

2 c˜j+1pj ∆ pj− = − (4.102) 1 m therefore we get the following diﬀerential equation for φ(τ)

d2φ(τ) c˜(τ) = − φ(τ) (4.103) dτ 2 m

102 Furthermore, φ(τ) satisﬁes the following boundary conditions

φ(t) = lim p0 =0 →0 2 dφ(τ) p1 − p0 |τ=t = lim = lim 2 − c˜1 − 1 = 1 (4.104) dτ →0 →0 m Recalling eq. (4.91) we have

f(t ,t) = lim pn−1 = φ(t ) (4.105) →0 Therefore, for quadratic lagrangian we get the general result m q,t|q, t = eiScl (4.106) 2πif(t,t) with f(t,t)=φ(t).

Examples:

The free particle.

We need only to evaluate f(t,t). The function φ(τ)satisﬁes

d2φ(τ) dφ(τ) =0,φ(t)=0, |τ t = 1 (4.107) dτ 2 dτ = with solution φ(τ)=τ − t (4.108) from which f(t,t)=t − t (4.109) in agreement with the direct calculation.

The harmonic oscillator.

In this case b =0andc = mω2. Therefore

2 d φ(τ) 2 dφ(τ) = −ω φ(τ),φ(t)=0, |τ t = 1 (4.110) dτ 2 dτ = from which 1 φ(τ)= sin ω(τ − t) (4.111) ω by putting T = t − t: 1 f(t,t)= sin ωT (4.112) ω

103 To evaluate the classical action we need to solve the classical equations of motion

mq¨ + mω2q = 0 (4.113) with boundary conditions q(t)=q, q(t)=q (4.114) The general solution is given by

q(τ)=A sin ω(τ − t)+B cos ω(τ − t) (4.115)

From the boundary conditions

q − q cos ωT A = ,B= q (4.116) sin ωT and substituting inside the lagrangian 1 1 1 L = mq˙2 − mω2q2 = mω2 (A2 − B2)cos2ω(τ − t) − 2AB sin 2ω(τ − t) 2 2 2 (4.117) Recalling the following integrals

t sin 2ωT t 1 − cos 2ωT dτ cos 2ω(τ − t)= , dτ sin 2ω(τ − t)= (4.118) t 2ω t 2ω we get 1 mω S = (q2 + q2)cosωT − 2qq (4.119) cl 2 sin ωT and i mω mω (q2 + q 2)cosωT − 2qq q,t|q, t = e2 sin ωT (4.120) 2πi sin ωT In the limit ω → 0 we get back correctly the free case.

4.6 Functional formalism

The most natural mathematical setting to deal with the path integral is the theory of functionals. The concept of functional is nothing but an extension of the concept of function. Recall that a function is simply a mapping from a manifold M to R1, that a law which associates a number to each point of the manifold. A real functional is deﬁned as an application from a space of functions (and therefore an ∞-dimensional space) to R1. There are many spaces of functions. For instance, the space of the square integrable real functions of a real variable. Therefore a real

104 functional associate a real number to a real function. If we denote the space of functions by L, a real functional is the mapping

F : L→R1 (4.121)

We will make use of the notation

F [η],F,F[·],η∈L (4.122) to denote the functional of deﬁned in eq. (4.121). Examples of functionals are •L=L1[−∞, +∞]= the space of the integrable functions in R1 with respect to the measure w(x)dx: +∞ F1[η]= dxw(x)η(x) (4.123) −∞

•L=L2[−∞, +∞]= space of the square integrable functions in R1: 1 − dxη2(x) 2 F2[η]=e (4.124)

•L=C= space of the continuous functions:

F3[η]=η(x0) (4.125)

F3 is the functional that associates to each continuous function its value at x0. The next step is to define the derivative of a functional. We need to understand how a functional varies when we vary the function. A simple way of understand this point is to consider some path on the definition manifold of the functions. For instance consider functions from R1 to R1, and take an interval (t, t)ofR1 made of many infinitesimal pieces, as we did for defining the path integral. Then we can approximate a function η(t) with a convenient limit of its discrete approximation (η1,η2, ···,ηn) obtained by evaluating the functions at the discrete points in which the interval has been divided. In this approximation we can think of the functional F as an application from Rn → R1.Thatis

F [η] → F (η1, ···,ηn) ≡ F (ηi),i=1, ···,n (4.126)

Then we can consider the variation of F (ηi) when we vary the quantities ηi n ∂F(ηi) F (ηi + δηi) − F (ηi)= δηj (4.127) ∂ηj j=1 Let us put 1 ∂F Ki = (4.128) ∂ηi

105 where is the inﬁnitesimal interval where we evaluate the discrete apoproximation to the function. Then, for → 0 n F (ηi + δηi) − F (ηi)= Kjδηj → Kδηdx (4.129) j=1 that is δF[η]= K(x)δη(x)dx (4.130)

Since δη(x) is the variation of the function evaluated at x, we will define the functional derivative as δF δF[η]= δη(x)dx (4.131) δη(x) In other words. to evaluate the functional derivative of a functional we first evaluate its infinitesimal variation and then we can get the functional derivative by comparison with the previous formula. Looking at the eq. (4.128) we get

δF[η] 1 ∂F(ηi) = lim δη(x) →0 ∂ηi xn − x1 ηi = η(xi),xi = x +(i − 1), = ,i=1, ···,n(4.132) 1 n − 1 let us now evaluate the functional derivatives of the previous functionals:

δF1[η]= w(x)δη(x)dx (4.133) giving δF [η] 1 = w(x) (4.134) δη(x) Then 1 − dxη2(x) 2 δF2[η]=e (− η(x)δη(x)) (4.135) and therefore δF [η] 2 = −η(x)F [η] (4.136) δη(x) 2

In order to evaluate the derivative of F3 let us write

F3[η]= η(x)δ(x − x0)dx (4.137)

In this way F3 looks formally as a functional of type F1,and δF [η] 3 = δ(x − x ) (4.138) δη(x) 0

106 Let us consider also 1 − dxdyK(x, y)η(x)η(y) 2 F4[η]=e (4.139) then 1 − dxdyK(x, y)η(x)η(y) 2 δF4[η]=e (− dxdyK(x, y)δη(x)η(y)) (4.140) where we have used the symmetry of the integral with respect to x ↔ y.Weobtain δF [η] 4 = − dyK(x, y)η(y)F [η] (4.141) δη(x) 4

A further example is the action functional. In fact the action is a functional of the path t S[q]= L(q, q˙)dt (4.142) t By varying it

t ∂L ∂L t ∂L d ∂L δS[q]= δq + δq˙ dt = − δqdt (4.143) t ∂q ∂q˙ t ∂q dt ∂q˙

Since we consider variations with respect to functions with ﬁxed boundary conditions, we have δq(t)=δq(t) = 0. We get

δS ∂L d ∂L = − (4.144) δq(t) ∂q(t) dt ∂q˙(t)

Starting from the functional derivative one can generalize the Taylor expansion to the functional case. This generalization is called the Volterra expansion of a functional ∞ k 1 δ F [η] F [η0 + η1]= dx1 ···dxk η1(x1) ···η1(xk) (4.145) k! δη(x1) ···δη(xk) k=0 η=η0 All these deﬁnitions are easily generalized to the case of η a function from Rm → Rn, or from a manifold M to a manifold N.

4.7 General properties of the path integral

In this Section we would like to stress some important properties of the path integral, namely

107 • Invariance of the functional measure under translations - It follows immediately from the deﬁnition that q(τ) → q(τ)+η(τ),η(t)=η(t)=0 p(τ) → p(τ)+π(τ), for any π(t),π(t) (4.146)

• Factorization of the path integral. From the very deﬁnition of the path it follows that in order to evaluate the amplitude for going from q at time t to q at time t we can ﬁrst go to q”toatimet ≤ t” ≤ t, sum up over all the paths fromq a q”andfromq”toq, and eventually integrating over all the possible intermediate points q” (see Fig. 4.7.1)).

q q" q'

Fig. 4.7.1 - The factorization of the path integral.

In equations we get

t t ” qt i Ldτ + i Ldτ q ,t|q, t = dµ(q(τ))dµ(p(τ))e t t” q,t t ” q”,t” i Ldτ = dq” dµ(q(τ))dµ(p(τ))e t q,t t q,t i Ldτ × dµ(q(τ))dµ(p(τ))e t” q”,t” = dq” q,t|q”,t” q”,t”|q, t (4.147) as it follows from the factorization properties of the measure dµ(q(τ)) = dq(τ)=dq” dq(τ) dq(τ) t<τ

108 dµ(p(τ)) = dp(τ)= dp(τ) dp(τ) (4.148) t≤τ≤t” t≤τ≤t” t”≤τ≤t

It is clear that the factorization property is equivalent to the completeness of the states. Using these properties it is possible to show that the wave function, as defined in (4.68) satisfies the Schrödinger equation. The next important property of the path integral is that it gives an easy way of evaluating expectation values. To this end we will consider expressions of the type (which will be called the average value of F )

q,t dµ(q(τ))dµ(p(τ))F [q(τ),p(τ)]eiS (4.149) q,t where F is an arbitrary functional of q, p and of their time derivatives. Let us start with q,t dµ(q(τ))dµ(p(τ))eiSq(t”),t≤ t” ≤ t (4.150) q,t We begin by the observation that

q,t q,t dµ(q(τ))dµ(p(τ))eiSq(t)=q dµ(q(τ))dµ(p(τ))eiS = q q,t|q, t q,t q,t (4.151) since q(t) satisﬁes the boundary conditions q(t)=q, q(t)=q.Byusingthe factorization we get q,t q”,t” dµ(q(τ))dµ(p(τ))eiSq(t”) = dq” dµ(q(τ))dµ(p(τ))eiSq(t”) q,t q,t q,t · dµ(q(τ))dµ(p(τ))eiS q”,t” = dq” q,t|q”,t” q” q”,t”|q, t (4.152)

Since |q, t is an eigenstate of q(t) it follows

q,t dµ(q(τ))dµ(p(τ))eiSq(t”) = dq” q,t|q(t”)|q”,t” q”,t”|q, t q,t = q,t|q(t”)|q, t (4.153)

Therefore the average value of q(t”) evaluated with exp(iS) coincides with the expectation value of the operator q(t”). Now let us study the average value of q(t1)q(t2)cont ≤ t1,t2 ≤ t . By proceeding as before we will get the expectation value of q(t1)q(t2)ift1 ≥ t2,orq(t2)q(t1)if

109 t2 ≥ t1.IntermsoftheT -product we get

q,t iS dµ(q(τ))dµ(p(τ))e q(t1)q(t2)= q ,t|T (q(t1)q(t2)|q, t (4.154) q,t Analogously

q,t iS dµ(q(τ))dµ(p(τ))e q(t1) ···q(tn)= q ,t|T (q(t1) ···q(tn))|q, t (4.155) q,t

Let us now consider a functional of q(t)such to admit an expansion in a Volterra’s series q,t dµ(q(τ))dµ(p(τ))F [q]eiS q,t ∞ k 1 δ F [q] = dt1 ···dtk k! δq(t1) ···δq(tk) k=0 q=0 q,t iS · dµ(q(τ))dµ(p(τ))e q(t1) ···q(tk) q,t ∞ k 1 δ F [q] = dt1 ···dtk q ,t|T (q(t1) ···q(tk))|q, t k! δq(t1) ···δq(tk) k=0 q=0 ≡ q,t|T (F [q])|q, t (4.156) where ∞ k 1 δ F [q] T (F [q]) ≡ dt1 ···dtk T (q(t1) ···q(tk)) (4.157) k! δq(t1) ···δq(tk) k=0 q=0 If the functional depends also on the time derivatives a certain caution is necessary. For instance, let us study the average value of q(t1)˙q(t2)(t ≤ t1,t2 ≤ t ):

q,t iS dµ(q(τ))dµ(p(τ))e q(t1)˙q(t2) q,t q,t − iS q(t2 + ) q(t2) = lim dµ(q(τ))dµ(p(τ))e q(t1) →0 q,t 1 = lim q,t|T q(t )q(t + ) − T q(t )q(t ) |q, t → 1 2 1 2 0 d = q ,t|T q(t1)q(t2) |q, t (4.158) dt2 Let us deﬁne an ordered T -product (or covariant T-product) as d T q(t1)q˙ (t2) = T q(t1)q(t2) (4.159) dt2

110 Then it is not diﬃcult to show that for an arbitrary functional of q(t)eq ˙(t) one has

q,t dµ(q(τ))dµ(p(τ))F [q, q˙]eiS = q,t|T (F [q, q˙ ])|q, t (4.160) q,t

The proof can be done by expanding F [q, q˙] in a Volterra’s series of q andq ˙,integrating by parts and using (4.156). At the end one has to integrate again by parts. The T product is obtained as in (4.158), The power of this formulation of quantum mechanics comes about when we want to get the fundamental properties of quantum mechanics as the equations of motion for the operators, the commutation relations and so on so forth. The fundamental relation follows from the following property of the path integral

q ,t δ dµ(q(τ))dµ(p(τ)) F [q(τ),p(τ)]eiS = 0 (4.161) q,t δq(τ)

This is a simple consequence of the deﬁnition of the path integral and of the functional derivative ⎡ ⎤ n−1 ⎢ ⎥ n−1 n−1 ⎢ i L(qk,pk)⎥ dpk 1 ∂ ⎢ k=0 ⎥ lim dqk ⎢F (q1, ···,qn−1)e ⎥ = 0 (4.162) n→∞ 2π ∂qk k=1 k=0 ⎣ ⎦

(of course we are assuming that the integrand goes to zero for large values of the qi’s. To this end one has to deﬁne the integral in an appropriate way as it will be shown later on). On the other hand the previous equation follows from the invariance under translation of the functional measure. In fact, let us introduce the following notation dµq,p = dµ(q(τ))dµ(p(τ)) (4.163) It follows for an inﬁnitesimal function η such that η(t)=η(t)=0

q ,t q ,t q ,t q ,t δA[q] dµq,p A[q]= dµq,p A[q+η]= dµq,p A[q]+ dµq,p η(t) dt q,t q,t q,t q,t δq(t) (4.164) Proving the validity of eq. (4.161). By performing explicitly the functional derivative in eq. (4.161)we get a relation which is the basis of the Schwinger’s formulation of quantum mechanics, that is δS δF[q] i q,t|T F [q] |q, t = − q,t|T |q, t (4.165) δq(τ) δq(τ) The importance of this relation follows from the fact that we can use this relation to get all the properties of quantum mechanics by selecting the functional F in a

111 convenient way. For instance, by choosing F [q]=1weget δS q,t|T |q, t = 0 (4.166) δq(t1) that is the quantum equations of motion. With the choice F [q]=q we get δS q ,t|T q(t1) |q, t = i q ,t|q, t δ(t1 − t2) (4.167) δq(t2) and since this relation holds for arbitrary states δS T q(t1) = iδ(t1 − t2) (4.168) δq(t2) For simplicity, let us consider the following action t 1 S = mq˙2 − V (q) dt (4.169) t 2 then δS ∂V (q(t2)) = − mq¨(t2)+ (4.170) δq(t2) ∂q(t2) from which ∂V (q) T q(t1) mq¨(t2)+ = −iδ(t1 − t2) (4.171) ∂q(t2) By using the deﬁnition of the T -product we get 2 d T (q(t1)q¨(t2)= 2 T (q(t1)q(t2)) dt2 d2 = [θ(t − t )q(t )q(t )+θ(t − t )q(t )q(t )] dt2 1 2 1 2 2 1 2 1 2 d = − δ(t1 − t2)q(t1)q(t2)+δ(t1 − t2)q(t2)q(t1) dt2

+ theta(t1 − t2)q(t1)q˙ (t2)+θ(t2 − t1)q˙ (t2)q(t1)

= δ(t1 − t2)q(t1)q˙ (t2)+δ(t1 − t2)q˙ (t2)q(t1)+T (q(t1)q¨(t2))

= δ(t1 − t2)[q(t1), q˙ (t1)] + T (q(t1)q¨(t2)) (4.172) Inserting this inside the eq. (4.171 ∂V −mδ(t1 − t2)[q(t1), q˙ (t2)] + T q(t1) mq¨(t2)+ = −iδ(t1 − t2) (4.173) ∂q(t2) and using the equations of motion, we get [q(t),mq˙ (t)] = i (4.174) Therefore we get the canonical commutator [q(t), p(t)] = i (4.175)

112 4.8 The generating functional of the Green’s functions

In ﬁeld theory the relevant amplitudes are the ones evaluated between t = −∞ and t =+∞. We have seen that using the reduction formulas we are lead to evaluate the vacuum expectation value of T -products of local operators. Therefore we will consider matrix elements of the type (limiting ourselves, for the moment being, to a single degree of freedom)

0|T (q(t1) ···q(tn))|0 (4.176)

Here |0 is the ground state of system. Notice that more generally one can consider the matrix element

q,t iS q ,t|T (q(t1) ···q(tn))|q, t = dµ(q(τ))dµ(p(τ))e q(t1) ···q(tn) (4.177) q,t

These matrix elements can be derived in a more compact form, by introducing the generating functional

q,t iSJ q ,t|q, t J = dµ(q(τ))dµ(p(τ))e (4.178) q,t where t SJ = dτ [pq˙ − H + Jq] (4.179) t and J(τ) is an arbitrary external source. By diﬀerentiating this expression with respect to the external source and evaluating the derivatives at J =0weget n n 1 δ q ,t|T (q(t1) ···q(tn))|q, t = q ,t|q, t J (4.180) i δJ(t1) ···δJ(tn) J=0 Now we want to see how it is possible to bring back the evaluation of the matrix element in (4.176) to the knowledge of the generating functional. Let us introduce the wave function of the ground state

113 Next we deﬁne the functional Z[J], the generating functional of the vacuum amplitudes, that is | | Z[J]= dq dqΦ0(q ,t) q ,t q, t J Φ0(q, t)= 0 0 J (4.183)

It follows n 1 δn 0|T (q(t1) ···q(tn))|0 = Z[J] (4.184) i δJ(t1) ···δJ(tn) J=0 We want to show that it is possible evaluate Z[J] as a path integral with arbitrary boundary conditions on q and q. That is we want to prove the following relation iE0(t − t) e Z[J] = lim q ,t|q, t J (4.185) t→ i∞,t→−i∞ + Φ0(q)Φ0(q ) wit q and q arbitrary and with the ground state wave functions taken at t =0.To this end let us consider an external source J(t) vanishing outside the interval (t,t) with t >t >t >t. Then, we can write q ,t|q, t J = dq dq q ,t|q ,t q ,t |q ,t J q ,t |q, t (4.186) with −iH(t − t ) −iEn(t − t ) q ,t|q ,t = q |e |q = Φn(q )Φn(q )e (4.187) n

Since E0 is the lowest eigenvalue we get iE0t iEnt −i(En − E0)t lim e q ,t|q ,t = lim Φn(q )Φn(q )e e = t→−i∞ t→−i∞ n iE0t =Φ0(q )Φ0(q )e =Φ0(q ,t )Φ0(q ) (4.188) Analogously −iEn(t − t) q ,t |q, t = Φn(q )Φn(q)e (4.189) n and −iE0t −iEnt i(En − E0)t lim e q ,t |q, t = lim Φn(q )Φn(q)e e = t→+i∞ t→+i∞ n −iE0t =Φ0(q )Φ0(q)e =Φ0(q ,t )Φ0(q) (4.190) And ﬁnally we get the result − iE0(t t) | im t→+i∞,t →−∞e q ,t q, t J | = dq dq Φ0(q)Φ0(q )Φ0(q ,t ) q ,t q ,t J Φ0(q ,t )= =Φ0(q)Φ0(q )Z[J] (4.191)

114 Let us notice again that the values q and q are completely arbitrary, and furthermore that the coeﬃcient in front of the matrix element is J-independent. Therefore it is physically irrelevant. In fact, the relevant physical quantities are the ratios n n 0|T (q(t ) ···q(tn))|0 1 1 δ 1 = Z[J] (4.192) 0|0 i Z[0] δJ(t1) ···δJ(tn) J=0 where the J-independent factor cancels out. Then we can write

t q,t i (L + Jq)dτ D t Z[J] = lim N (q(τ))e (4.193) t→+i∞,t →−i∞ q,t where N is a J-independent normalization factor. This expression suggests the deﬁ- nition of an euclidean generating functional ZE[J]. This is obtained by introducing an euclidean time τ = it (Wick’s rotation)

τ dq q,τ (L(q, i )+Jq)dτ E D τ dτ Z [J] = lim N (q(τ))e (4.194) τ →+∞,τ→−∞ q,τ

Consider for instance the case 1 dq L = m( )2 − V (q) (4.195) 2 dt from which dq 1 dq 2 L(q, i )=− m − V (q) (4.196) dτ 2 dτ It is then convenient to deﬁne the euclidean lagrangian and euclidean action 1 dq 2 LE = m + V (q) (4.197) 2 dτ

τ SE = LEdτ (4.198) τ Notice that the euclidean lagrangian coincides formally with the hamiltonian. More generally dq LE = −L(q, i ) (4.199) dτ Therefore

τ q ,τ −SE + Jqdτ E D τ Z [J] = lim N (q(τ))e (4.200) τ →+∞,τ→−∞ q,τ

115 − Since e SE is a positive definite functional, the path integral turns out to be well defined and convergent if SE is positive definite.the vacuum expectation values of the operators q(t) can be evaluated by using ZE and continuing the result for real times n n 1 δ Z[J] 1 δ ZE[J] =(i)n (4.201) Z[0] δJ(t1) ···δJ(tn) J=0 ZE[0] δJ(τ1) ···δJ(τn) J=0,τi=iti Since the values of q and q are arbitrary a standard choice is to set them at zero.

4.9 The Green’s functions for the harmonic oscillator

It is interesting to evaluate the generating functional for the harmonic oscillator starting from both the equations (4.185) and (4.200). Let us start from eq. (4.183), where we do not need to specify the boundary conditions D iSJ Z[J]= (q(τ))Φ0(q ,t)e Φ0(q, t)dqdq (4.202) with 1 1 2 i mω 4 − mωq − ωt Φ (q, t)= e 2 e 2 (4.203) 0 π

1 1 2 i mω 4 − mωq ωt Φ (q,t)= e 2 e2 (4.204) 0 π and t 1 2 1 2 2 SJ = mq˙ − mω q + Jq dτ (4.205) t 2 2 In this calculation we are not interested in the normalization factors but only on the J dependence. Then, let us consider the argument of the exponential in eq. (4.9 and, for the moment being, let us neglect the time dependent terms from the oscillator wave functions 1 t 1 1 arg = − mω(q2 + q2)+i mq˙2 − mω2q2 + Jq dτ (4.206) 2 t 2 2 let us perform the change of variable

q(τ)=x(τ)+x0(τ) (4.207) where we will try to arrange x0(τ) in such a way to extract all the dependence from the source J out of the integral. We have 1 1 arg = − mω[x(t)2 + x2(t)] − mω[x (t)2 + x2(t)] − mω[x(t)x (t)+x(t)x (t)] 2 2 0 0 0 0

116 t 1 1 t 1 1 + i mx˙ 2 − mω2x2 dτ + i mx˙ 2 − mω2x2 + Jx dτ 2 2 2 0 2 0 0 t t t 2 + i mx˙x˙ 0 − mω xx0 + Jx dτ (4.208) t Integrating by parts the last integral can be written as

t t 2 im[xx˙ 0]t − i [mx¨0 + mω x0 − J]xdτ (4.209) t and choosing x0 as a solution of the wave equation in presence of the source J 1 x¨ + ω2x = J (4.210) 0 0 m we get im[x(t )˙x0(t ) − x(t)˙x0(t)] (4.211)

Let us take the following boundary conditions on x0:

ix˙ 0(t )=ωx0(t ),ix˙ 0(t)=−ωx0(t) (4.212)

In this way, the term (4.211) cancels the mixed terms 1 1 arg = − mω[x2(t)+x2(t)] − mω[x2(t)+x2(t)] 2 2 0 0 t 1 1 + i [ mx˙ 2 − mω2x2]dτ t 2 2 t 1 2 − 1 2 2 + i [ mx˙ 0 mω x0 + Jx0]dτ (4.213) t 2 2 Integrating by parts the last term we get

t 1 2 − 1 2 2 i [ mx˙ 0 mω x0 + Jx0]dτ = t 2 2 t 1 2 1 1 i t = [− mx0(¨x0 + ω x0 − J)+ Jx0]dτ + m[x0x˙ 0]t = t 2 m 2 2 t 1 2 2 i = mω[x0(t )+x0(t)] + Jx0dτ (4.214) 2 2 t

Keeping in mind the boundary conditions (4.212), the term coming from the integration by parts cancels the second term in eq. (4.213), and therefore

t t 1 2 2 1 2 1 2 2 i arg = − mω[x (t)+x (t )] + i [ mx˙ − mω x ]dτ + Jx0dτ (4.215) 2 t 2 2 2 t

117 The J dependence is now in the last term, and we can extract it from the integral. The ﬁrst two term give rise to the functional at J =0.Weobtain

i t J(τ)x0(τ)dτ Z[J]=e2 t Z[0] (4.216) with x0(τ) solution of the equations of motion (4.210) with boundary conditions (4.212). For t<τ

x0(τ)=i ∆(τ − s)J(s)ds (4.217) t with d2 i + ω2 ∆(τ − s)=− δ(τ − s) (4.218) dτ 2 m and i∆(˙ t − s)=ω∆(t − s),i∆(˙ t − s)=−ω∆(t − s) (4.219) These equations are easily solved fort t >sand t~~0 ∆(τ)=Be+iωτ ,τ<0 (4.220)~~

~~Therefore we have − ∆(τ)=Aθ(τ)e iωτ + Bθ(−τ)e+iωτ (4.221)~~

The constants A e B are fixed by the requirement that ∆(τ) satisfies eq. (4.218). We have − ∆(˙ τ)=(A − B)δ(τ) − iAωθ(τ)e iωτ + iBωθ(−τ)e+iωτ (4.222) and ∆(¨ τ)=(A − B)δ˙(τ) − iω(A + B)δ(τ) − ω2∆(τ) (4.223) from which 1 A = B, A = (4.224) 2ωm and finally 1 − ∆(τ)= θ(τ)e iωτ + θ(−τ)e+iωτ (4.225) 2ωm The functional we were looking for is

~~1 t − dsdsJ(s)∆(s − s)J(s) Z[J]=e 2 t Z[0] (4.226)~~

118 We see that ∆(s − s) is the vacuum expectation value of the T -product of the position operators at the times s and s: 1 δ2Z[J] 0|T (q(s)q(s))|0 − − ∆(s s )= = (4.227) Z[0] δJ(s)δJ(s ) J=0 0|0

~~The analogue calculation in the euclidean version is simpler since the path integral is of the type already considered (quadratic action)~~

~~cl −S( ) ZE[J] ≈ e E (4.228)~~

(cl) Here SE is th euclidean classical action. We have already noticed that here the boundary conditions are arbitrary and therefore we choose q(τ) → 0forτ →±∞. we have to solve the equations of motion in presence of the external source +∞ δ 2 SE − Jqdτ = −mq¨(τ)+mω q(τ) − J = 0 (4.229) δq(τ) −∞ that is 1 q¨ − ω2q = − J (4.230) m Let us now deﬁne the euclidean Green’s function 2 d 2 1 − ω DE(τ)=− δ(τ) (4.231) dτ 2 m with the boundary condition lim DE(τ) = 0 (4.232) τ→±∞ then, the solution of the eq. (4.230) is +∞ q(τ)= DE(τ − s)J(s)ds (4.233) −∞

~~We can evaluate DE(τ) starting from the Fourier transform −iντ DE(τ)= dνe DE(ν) (4.234)~~

~~Substituting inside (4.231) we get −iντ 2 2 1 − dνe [ν + ω ]DE(ν)=− δ(τ) (4.235) m from which 1 1 DE(ν)= (4.236) 2πm ν2 + ω2~~

119 and +∞ 1 1 −iντ DE τ dν e ( )= 2 2 (4.237) 2πm −∞ ν + ω To evaluate this integral we close the integral with the circle at the ∞ in the lower ν-plane for τ>0 or in the upper ν-plane for τ<0. The result is −ωτ 1 e 1 −ωτ DE(τ)= (−2πi)= e (4.238) 2πm (−2iω) 2mω

~~By doing the same calculation for τ<0 we ﬁnally get~~

1 −ω|τ| DE(τ)= e (4.239) 2mω The classical action is given by +∞ +∞ +∞ 1 2 1 1 1 SE − Jqdτ = − m q¨ − ω q + J q − Jq dτ = − Jqdτ −∞ −∞ 2 m 2 2 −∞ (4.240) and therefore +∞ +∞ 1 1 Jqdτ J(s)DE(s − s )J(s )dsds 2 2 ZE[J]=e −∞ Z[0] = e −∞ Z[0] (4.241)

~~It follows 2 1 δ ZE[J] − = DE(s s ) (4.242) ZE[0] δJ(s)δJ(s ) J=0 and using the eq. (4.201)~~

2 2 1 δ Z[J] 1 δ ZE[J] = − (4.243) Z[0] δJ(t1)δJ(t2) J=0 ZE[0] δJ(τ1)δJ(τ2) J=0,τi=iti we get ∆(t)=DE(it) (4.244) To compare these two expressions let us introduce the Fourier transform of ∆(t). Since it satisﬁes eq. (4.218) we obtain − ∆(t)= dνe iνt∆(ν) (4.245) with i (−ν2 + ω2)∆(ν)=− (4.246) 2πm that is i 1 ∆(ν)= (4.247) 2πm ν2 − ω2

120 Whereas eq. (4.237) does not present any problem (ν2 + ω2 > 0), in eq. (4.246) we need to deﬁne the singularities present in the denominator. In fact, this expression hastwopolesatν = ±ω, and therefore we need to specify how the integral is deﬁned. In order to reproduce the euclidean result, it is necessary to specify the integration path as in Fig. 4.9.1. For instance, for t>0 by closing the integral in thelowerplaneweget i 1 − 1 − (−2πi) e iωt = e iωt (4.248) 2πm 2ω 2mω

~~. Im ν~~

~~− ω + ω Re ν~~

~~Fig. 4.9.1 - The integration path for equation (4.245).~~

We would get the same result by integrating along the real axis but translating the denominator of +i (>0) in such a way to move the pole at ω in the lower plane and the one at −ω in the upper plane. We get

~~∞ − i + e iνt ∆(t) = lim dν (4.249) + 2 2 →0 2πm −∞ ν − ω + i~~

~~With this prescription we can safely rotate the integration path anti-clockwise by 900, since we do not encounter any singularity (see Fig. 4.9.2)~~

~~i∞ − i + e iνt t dν ∆( )= 2 2 (4.250) 2πm −i∞ ν − ω~~

We have omitted here the limit since the expression is now regular. Performing the change of variable ν = iν +∞ +ν t 1 e ∆(t)=− dν = DE(it) (4.251) 2 2 2πm −∞ −ν − ω Therefore we have shown how it works the analytic continuation implicit in eq. (4.244). By doing so we have also seen that the Feynman prescription (the +i term) is equivalent to the euclidean formulation. One could arrive to this connection also

~~121 . Im ν~~

~~− ω + iε~~

~~ν ω − i ε Re~~

Fig. 4.9.2 - The Wick rotation in the ν-plane. by noticing that the Feynman prescription is equivalent to substitute in the path integral the term i − mω2 q2(t)dt e 2 (4.252) with i − m(ω2 − i) q2(t)dt e 2 (4.253) Therefore the Feynman prescription is there to make the integral convergent. 1 − m q2(t)dt e 2 (4.254)

The same convergence is ensured by the euclidean formulation. Therefore we have shown that, in the case of quadratic actions, the path integral can be formulated as an euclidean integral, making it a well deﬁned mathematical expression. This fact is enough to guarantee that all the manipulations that we make with the path integral in the perturbative expansion are well deﬁned, since we will be expanding around the quadratic case.

~~122 Chapter 5~~

~~The path integral in ﬁeld theory~~

~~5.1 The path integral for a free scalar ﬁeld~~

We will extend now the previous approach to the case of a scalar field theory. We will show later how to extend the formulation to the case of Fermi fields. Let us start with a free scalar field. We have denoted the quantum operator by φ(x). Here we will denote the classical field by ϕ(x). For a neutral particle this is a real function, and its dynamics is described by the lagrangian t 4 1 µ 2 2 3 1 µ 2 2 S = d x ∂µϕ∂ ϕ − m ϕ ≡ dt d x ∂µϕ∂ ϕ − m ϕ (5.1) V 2 t 2 A free field theory can be seen as a continuous collection of non-interacting harmonic oscillators. As it is well known this can be seen by looking for the normal modes. To this end, let us consider the Fourier transform of the field 1 · ϕ(x, t)= d3keik xq(k, t) (5.2) (2π)3 By substituting inside eq. (5.1) we get 3 3 1 4 d k d k S = d x 3 3 [˙q(k, t)˙q(k ,t) 2 V (2π) (2π) · · +(k · k − m2)q(k, t)q(k,t)]eik x + ik x 1 t d3k dt q k, t q −k,t − |k|2 m2 q k, t q −k, t = 3 [˙( )˙( ) ( + ) ( ) ( )] (5.3) 2 t (2π) Since ϕ(x, t)isarealfield q∗(k, t)=q(−k,t) (5.4) and t d3k 1 | |2 − 2| |2 2 ||2 2 S = dt 3 q˙(k, t) ωk q(k, t) ,ωk = k + m (5.5) t (2π) 2

123 Therefore the system is equivalent to a continuous set of complex oscillators q(k, t) (or a pair of real oscillators)). The equations of motion are 2 q¨(k, t)+ωkq(k, t) = 0 (5.6) We are now in the position of evaluating the generating functional Z[J] by adding an external source term to the action ! 4 1 µ 2 2 S = d x ∂µϕ∂ ϕ − m ϕ + Jϕ (5.7) V 2 From the Fourier decomposition we get ! t d3k 1 | |2 − 2| |2 − S = dt 3 q˙(k, t) ωk q(k, t) + J( k, t)q(k, t) (5.8) t (2π) 2 wherewehavealsoexpandedJ(x) in terms of its Fourier components. Let us separate q(k, t)andJ(k,t) into their real and imaginary parts

~~q(k, t)=x(k,t)+iy(k, t) J(k,t)=l(k, t)+im(k,t) (5.9)~~

It follows from the reality conditions that the real parts are even functions of k whereas the imaginary parts are odd functions x(k, t)=x(−k,t),l(k, t)=l(−k, t) y(k, t)=−y(−k, t),m(k, t)=−m(−k,t) (5.10) Now we can use the result found in the previous Section for a single oscillator. it will be enough to take the mass equal to one and sum over all the degrees of freedom. In particular we get for the exponent in eq. (4.226)

~~1 t − dsdsJ(s)∆(s − s)J(s) → 2 t t 3 1 d k →− − dsds 3 l(k,s)∆(s s ; ωk)l(k, s ) 2 t (2π) + m(k, s)∆(s − s ; ωk)m(k, s )~~

t 3 1 d k − dsds J −k,s s − s ω J k, s = 3 ( )∆( ; k) ( ) (5.11) 2 t (2π) where we have made use of eq. (5.9). Going back to J(x, t), we get 3 1 4 4 d k ik · (x − y) − − d xd yJ(x) 3 ∆(s s ; ωk)e J(y) 2 (2π) i 4 4 2 ≡− d xd yJ(x)∆F (x − y; m )J(y) (5.12) 2

124 wherewehavedeﬁnedxµ =(x, s), yµ =(y, s)and 3 4 2 d k ik · x d k i −ikx i∆F (x, m )= ∆(s; ωk)e = lim e (5.13) (2π)3 →0+ (2π)4 k2 − m2 + i Again we have put kµ =(ν,k). Therefore, the generating functional is i 4 4 2 − d xd yJ(x)∆F (x − y; m )J(y) Z[J]=e 2 Z[0] (5.14) In particular we get 2 1 1 δ Z[J] 2 0|T (φ(x1)φ(x2))|0 = − = i∆F (x1 − x2; m ) (5.15) 0|0 Z[0] δJ(x1)δJ(x2) J=0 We could repeat the exercise in the euclidean formulation, with the result 1 4 4 2 d xd yJ(x)G0(x − y; m )J(y) ZE[J]=e2 ZE[0] (5.16) where d4k 1 − G (x; m2)= e ikx (5.17) 0 (2π)4 k2 + m2 We go from Minkowskian to the Euclidean description through the substitution

~~−i∆F → G0 (5.18) For the future it will be convenient to use the following notation for the N-points Green’s functions | ··· | (N) 0 T (φ(x1) φ(xN )) 0 G (x , ···,xN )= (5.19) 1 0|0~~

(N) (N) In the free case we can get evaluate easily the generic G In fact G0 (the index recalls that we are in the free case) is obtained by developing the generating functional ∞ n Z[J] (i) 0|T (φ(x ) ···φ(xn))|0 4 ··· 4 ··· 1 ≡ = d x1 d xnJ(x1) J(xn) | Z[0] n n! 0 0 =0 ∞ n 4 4 (i) (n) ≡ d x ···d xnJ(x ) ···J(xn) G (x , ···,xn) (5.20) 1 1 n! 0 1 n=0 By comparison with the expansion of eq. (5.14) Z[J] i 4 4 =1− d x1d x2J(x1)J(x2)∆F (x1 − x2) Z[0] 2 1 4 4 − d x1d x2J(x1)J(x2)∆F (x1 − x2) 8 4 4 · d x3d x4J(x3)J(x4)∆F (x3 − x4)+··· (5.21)

~~125 Since Z[J]iseveninJ we have (2k+1) ··· G0 (x1, ,x2k+1) = 0 (5.22)~~

(2) (4) For G0 we get back eq. (5.13). To evaluate G0 we have ﬁrst to notice that (N) all the G0 ’s are symmetric in their arguments. therefore in order to extract the coeﬃcients we need to symmetrize. Therefore we write 4 4 d x1 ···d x4J(x1) ···J(x4)∆F (x1 − x2)∆F (x3 − x4) 1 4 4 = d x1 ···d x4J(x1) ···J(x4) ∆F (x1 − x2)∆F (x3 − x4) 3

~~+∆F (x1 − x3)∆F (x2 − x4)+∆F (x1 − x4)∆F (x2 − x3) (5.23)~~

~~It follows (4) G (x1, ···,x4)=− ∆F (x1 − x2)∆F (x3 − x4)+∆F (x1 − x3)∆F (x2 − x4) 0~~

~~+∆F (x1 − x4)∆F (x2 − x3) (5.24)~~

(2) (4) If we represent G0 (x, y) by a line as in Fig. 5.1.1, G0 will be given by Fig. (2) (2n) 5.1.2 since it is obtained in terms of products of G0 . Therefore G0 is given by a combination of products of n two point functions. We will call disconnected a diagram in which we can isolate two subdiagrams with no connecting lines. We see that in the free case all the non vanishing Green’s functions are disconnected except for the two point function. Therefore it is natural to introduce a functional generating only the connected Green’s functions. In the free case we can put . x y (2) Fig. 5.1.1 - The two point Green’s function G0 (x, y).

~~x1 x x x x1 x2 2 1 2~~

~~+ +~~

~~x3 x4 x3 x4 x3 x4 (4) Fig. 5.5.2 - The expansion of the four point Green’s function G0 (x1,x2,x3,x4).~~

~~Z[J]=eiW [J] (5.25)~~

126 with 1 4 4 W [J]=− d x d x J(x )J(x )∆F (x − x )+W [0] (5.26) 2 1 2 1 2 1 2 and we have 1 δ2Z[J] δ2W [J] = i = −i∆F (x1 − x2) (5.27) Z[0] δJ(x1)δJ(x2) J=0 δJ(x1)δJ(x2) J=0 Therefore the derivatives of W [J] give the connected diagrams with the proper normalization(that is divided by 1/Z[0]). In the next Section we will see that this property generalizes to the interacting case. That is W [J] is deﬁned through eq. (5.25), and its Volterra expansion gives rise to the connected Green’s functions ∞ n (i) 4 4 (n) iW [J]= d x ···d xnJ(x ) ···J(xn)G (x , ···,xn) (5.28) n! 1 1 c 1 n=0 where (n) 1 G (x , ···,xn)= 0|T (φ(x ) ···φ(xn))|0 (5.29) c 1 0|0 1 conn The index ”conn” denotes the connected Green’s functions. Notice that once we ﬁnd the connected Green’s functions the theory is completely solved, since the generating functional is recovered by exponentiation of W [J].

~~5.2 The generating functional of the connected Green’s functions~~

~~Let us start by deﬁning recursively the connected Green’s functions.~~

(1) (1) (1) G (x1)=Gc (x1) ≡ Gc ; (2) (2) (1) (1) (2) (1)2 G (x1,x2)=Gc (x1,x2)+Gc (x1)Gc (x2) ≡ Gc + Gc ; (3) (3) (1) (2) G (x1,x2,x3)=Gc (x1,x2,x3)+Gc (x1)Gc (x2,x3) (1) (2) (1) (2) (1) (1) (1) +Gc (x2)Gc (x1,x3)+Gc (x3)Gc (x1,x2)+Gc (x1)Gc (x2)Gc (x3) ≡ (3) (1) (2) (1)3 ≡ Gc +3Gc Gc + Gc (5.30)

On the right hand side we have used a symbolic notation by adding together the (n) topological equivalent conﬁgurations. To deﬁne Gc (x1, ···,xn) we decompose the set (x1, ···,xn) in all the possible subsets, and then we write (n) (n1) (n2) (nk) ··· n p ··· p q ··· q ··· r ··· r G (x1, ,x )= Gc (x 1 , ,x n1 )Gc (x 1 , ,x n2 ) Gc (x 1 , ,x nk ) (5.31) where we sum over all the indices ni such that n1 + ···+ nk = n and over all the ··· permutations from 1 to n of the indices in the set p1, ,pn1 , etc. A graphical

~~127 .~~

~~xxxxx~~

~~xxxxx xxxxx = c xxxxx~~

~~xxxxx~~

~~xxxxx c~~

~~xxxxx xxxxx c + xxxxx = xxxxx c~~

~~c c~~

~~xxxxx~~

~~xxxxx c xxxxx = c + 3 +~~

~~xxxxx c c~~

~~c c xxxxxx~~

~~xxxxxx xxxxxx c xxxxxx = + 3 + 4 + xxxxxx c c~~

c c c + 6 c + c c c Fig. 5.2.1 - The connected Green’s functions. description is given in Fig. 5.2.1. By using these definitions it is not difficult to check that W [J] is the generating functional of the connected Green’s functions. It (n) is enough to compare the expansion of Z[J]inseriesofGc with the expansion from the G(n). This can be done by induction. We will limit ourselves to an explicit check up to the 4th order. By defining G(0) = 1 and using the same symbolic notation as before, we get

∞ n Z[J] (i) n n − = J G( ) = ei(W [J] W [0]) Z[0] n! n=0 i2 i3 i4 =expiG(1)J + G(2)J 2 + G(3)J 3 + G(4)J 4 + ··· c 2! c 3! c 4! c 2 3 4 (1) i (2) 2 i (3) 3 i (4) 4 ≈ 1+iGc J + Gc J + Gc J + Gc J 2! 3! 4! 4 4 1 2 (1)2 2 i (2)2 4 3 (1) (2) 3 i (1) (3) 4 + i Gc J + Gc J + i Gc Gc J + Gc Gc J 2! 4! 3 1 3 3 2 + i3G(1) J 3 + i4G(1) G(2)J 4 3! c 2 c c 1 4 + i4G(1) J 4 + ··· 4! c

128 2 i 2 =1+iJG(1) + G(2) + G(1) c 2! c c 3 i 3 + J 3 G(3) +3G(1)G(2) + G(1) 3! c c c c 4 i 2 2 4 + J 4 G(4) +3G(2) +4G(1)G(3) +6G(1) G(2) + G(1) + ··· (5.32) 4! c c c c c c c 5.3 The perturbative expansion for the theory λϕ4

Let us consider an interacting scalar ﬁeld. The action will be of the type 4 1 µ 2 2 S = d x ∂µϕ∂ ϕ − m ϕ − V (ϕ) (5.33) V 2 where V (ϕ) is the interaction potential. A typical example is the theory λϕ4 with the potential given by λ V (ϕ)= ϕ4 (5.34) 4! The derivation of the perturbative expansion is very simple in this formalism. We take advantage of the following identity valid for any functional F iS + i d4xJ(x)ϕ(x) D(ϕ(x))F [ϕ]e 4 1 δ iS + i d xJ(x)ϕ(x) = F D(ϕ(x))e (5.35) i δJ

We separate in eq. (5.33) the quadratic part of the action, S0, from the interacting part and then we use the previous identity iS + i d4xJ(x)ϕ(x) Z[J]=N D(ϕ(x))e 4 4 −i d xV (ϕ) iS0 + i d xJ(x)ϕ(x) = N D(ϕ(x))e e 1 δ −i d4xV i δJ = e Z0[J] (5.36) Notice that we have suppressed the temporal limits in the expression for the generating functional. Z0[J] is the generating functional for the free case, evaluated in eq. (5.14). Therefore we get 4 1 δ i 4 4 −i d xV − d xd yJ(x)∆F (x − y)J(y) Z[J]=eiW [J] = Z[0]e i δJ e 2 (5.37)

129 In this expression we have suppressed the mass in the argument of ∆F .Forthe following calculation it will be useful to introduce the following shorthand notation 4 4 d x1 ···d xnF (x1, ···,xn) ≡ F (x1, ···,xn) (5.38)

~~Then, eq. (5.37) can be rewritten in the following way − 1 δ i i V − J(1)∆F (1, 2)J(2) iW [J] i δJ 2 Z[J]=e = e e Z[0] 1 δ −i V = e i δJ eiW0[J] (5.39)~~

This gives rise to the following expression for the generating functional of the connected Green’s functions ⎡ ⎛ ⎞ ⎤ − 1 δ ⎢ ⎜ i V ⎟ ⎥ W [J]=−i log ⎣eiW0[J] + ⎝e i δJ − 1⎠ eiW0[J]⎦

~~⎡ ⎛ ⎞ ⎤ − 1 δ ⎢ ⎜ i V ⎟ ⎥ −iW0[J] i δJ iW0[J] = W0[J] − i log ⎣1+e ⎝e − 1⎠ e ⎦(5.40)~~

This expression can be easily expanded in V . In fact, by deﬁning ⎛ ⎞ − 1 δ − ⎜ i V ⎟ δ[J]=e iW0[J] ⎝e i δJ − 1⎠ eiW0[J] (5.41) we see that we can expand W [J]inaseriesofδ. Notice also that the expansion is in term of the interaction lagrangian. At the second order in δ we get 1 W [J]=W [J] − i δ − δ2 + ··· (5.42) 0 2

~~Let us now consider the case of λϕ4. the functional δ[J] can be in turn expanded in a series of the dimensionless coupling λ~~

~~2 δ = λδ1 + λ δ2 + ··· (5.43)~~

~~It follows 1 W [J]=W [J] − iλδ − iλ2 δ − δ2 + ··· (5.44) 0 1 2 2 1~~

130 Then, from eq. (5.43) we get i − 1 δ 4 δ = − e iW0[J] eiW0[J] 1 4! i δJ 1 − 1 δ 4 1 δ 4 δ = − e iW0[J] eiW0[J] (5.45) 2 2(4!)2 i δJ i δJ By using the following list of functional derivatives δ eiW0[J] = −i∆(x, 1)J(1)eiW0[J] δJ(x) δ2 eiW0[J] =(−i∆(x, x) − ∆(x, 1)∆(x, 2)J(1)J(2)) eiW0[J] δJ(x)2 δ3 eiW0[J] − x, x x, J 3 = 3∆( )∆( 1) (1) δJ(x) + i∆(x, 1)∆(x, 2)∆(x, 3)J(1)J(2)J(3) eiW0[J] δ4 eiW0[J] =(−3∆(x, x)2 +6i∆(x, x)∆(x, 1)∆(x, 2)J(1)J(2) δJ(x)4 +∆(x, 1)∆(x, 2)∆(x, 3)∆(x, 4)J(1)J(2)J(3)J(4))eiW0[J] (5.46) where the integration on the variables 1, 2, 3, 4 is understood, we obtain i δ1 = − ∆(y,1)∆(y,2)∆(y,3)∆(y,4)J(1)J(2)J(3)J(4) 4! +6i ∆(y,y)∆(y,1)∆(y,2)J(1)J(2) −3 ∆2(y,y) (5.47)

The quantity δ2 canbemadeintermsofδ1 by writing 1 − 1 δ 4 − 1 δ 4 δ = − e iW0[J] eiW0[J]e iW0[J] eiW0[J] 2 2(4!)2 i δJ i δJ i − 1 δ 4 = − e iW0[J] eiW0[J]δ (5.48) 2(4!) i δJ 1 Then we have 1 δ 4 δ4eiW0[J] δ3eiW0[J] δδ eiW0[J]δ = δ +4 1 i δJ 1 δJ(x)4 1 δJ(x)3 δJ(x) δ2eiW0[J] δ2δ δeiW0[J] δ3δ +6 1 +4 1 δJ(x)2 δJ(x)2 δJ(x) δJ(x)3 δ4δ + eiW0[J] 1 (5.49) δJ(x)4

131 and therefore 1 i − δ3eiW0[J] δδ δ = δ2 − e iW0[J] 4 1 2 2 1 2 · 4! δJ(x)3 δJ(x) δ2eiW0[J] δ2δ δeiW0[J] δ3δ +6 1 +4 1 δJ(x)2 δJ(x)2 δJ(x) δJ(x)3 δ4δ + eiW0[J] 1 (5.50) δJ(x)4

2 The δ1 gives a disconnected contribution since there are no propagators connecting the two terms. In fact the previous expression shows that δ2 contains a term which 2 cancels precisely the term δ1 in eq. (5.44). By using the expression for δ1 and eqs. (5.46) we get 1 − 2 δ2 δ1 = 2 − i − i − = · 4 3∆(x, x)∆(x, 1)J(1) 2 4! 4! + i∆(x, 1)∆(x, 2)∆(x, 3)J(1)J(2)J(3) · 4∆(y,x)∆(y,4)∆(y,5)∆(y,6)J(4)J(5)J(6) + 12i∆(y,y)∆(y,x)∆(y,4)J(4) +6− i∆(x, x) − ∆(x, 1)∆(x, 2)J(1)J(2) 12∆2(y,x)∆(y,3)∆(y,4)J(3)J(4) +12i∆(y,y)∆2(y,x) +4− i∆(x, 1)J(1) 24∆3(y,x)∆(y,2)J(2) + 24∆4(y,x) (5.51)

~~By omitting the terms independent on the external source we have~~

− 1 2 δ2 δ1 = 2 i 1 1 = J(1)∆(1,x) ∆3(x, y)+ ∆(x, x)∆(x, y)∆(y,y) ∆(y,2)J(2) 2 6 4 i + J(1)∆(1,x)∆2(x, y)∆(y,y)∆(x, 2)J(2) 8 1 + J(1)∆(1,x)∆(x, x)∆(x, y)∆(y,2)∆(y,3)∆(y,4)J(2)J(3)J(4) 12 1 + J(1)J(2)∆(1,x)∆(2,x)∆2(x, y)∆(y,3)∆(y,4)J(3)J(4) 16 i − J(1)J(2)J(3)∆(1,x)∆(2,x)∆(3,x)∆(x, y)∆(y,4) 72 · ∆(y,5)∆(y,6)J(4)J(5)J(6) (5.52)

~~132 Now, by diﬀerentiating functionally eq. (5.44) and putting J =0wecanevaluate the connected Green’s functions. For instance, recalling that~~

2 (2) δ W [J] iGc (x1,x2)= (5.53) δJ(x1)δJ(x2) J=0 we get (2) λ 4 G (x ,x )=i∆F (x − x ) − d x∆F (x − x)∆F (x − x)∆F (x − x ) c 1 2 1 2 2 1 2 2 4 4 1 3 − iλ d xd y∆F (x − x) ∆ (x − y) 1 6 F 1 + ∆F (x − y)∆F (x − x)∆F (y − y) ∆F (y − x2) 4 2 λ 4 4 2 − i d xd y∆F (x − x)∆ (x − y)∆F (y − y)∆F (x − x ) (5.54) 4 1 F 2

~~xxxxxxxx~~

~~xx xxxxxxxx xx x x x xx xx~~

~~xx xxxxxxxx xx = x x + x xx xx +~~

~~xxxxxxxx x xxxxxxxx x x x x x x 1 xxxxxxxx 2 1 2 1 2~~

~~xx y x y~~

~~x xx xx xx xx x xx xx x xx xx + x xx xx xx + xx x xx + xx x xx xx x1 x2 x1 x x2 x1 x y x2 Fig. 5.3.1 - The graphical expansion for the two point connected Green’s function (2) Gc (x, y).~~

In Fig. 5.3.1) we give a graphical form to this expression. as we see the result has an easy interpretation. The diagrams of Fig. 5.3.1) are given in terms of two elements, the propagator ∆F (x − y), corresponding to the lines and the vertex (n) λϕ4 corresponding to a point joining four lines. In order to get Gc we have to draw all the possible connected diagrams containing a number of interaction vertices corresponding to the ﬁxed perturbative order. The analytic expression is obtained associating a factor i∆F (x − y) to each line connecting the point x with the point y, and a factor (−iλ/4!) for each interaction vertex. The numerical factors appearing in eq. (5.54) can be obtained by counting the possible ways to draw a given diagram. For instance let us consider the diagram of Fig. 5.3.2. This diagram contains the two external propagators, the internal one, and the vertex at x. To get the numerical factor we count the number of ways to attach the external propagators to the vertex, after that we have only a possibility to attach the vertex at the internal propagator. There are four ways to attach the the vertex at the ﬁrst propagator and three for the second. Therefore 1 1 · 4 · 3= (5.55) 4! 2

~~133 .~~

~~xx x~~

xx xx xx x xx xx xx xx xx xx = x xx xx xx x1 x x2 x1 x xx2 Fig. 5.3.2 - How to get the numerical factor for the ﬁrst order contributions to (2) Gc (x1,x2). where the factor 1/4! comes from the vertex. We proceed in the same way for the diagram of Fig. 5.3.3. .

~~x y x x y y~~

~~xx xx xx xx x x xx xx~~

~~xx xx xx xx = x x xx xx x1 x2 x1 x2 Fig. 5.3.3 - How to get the numerical factor for one of the second order contribu- (2) tions to Gc (x1,x2).~~

~~xx x xx xx x xx xx x xx xx 1 xxxxxxx 4 x1 xx xx 4 x xx xx x4 xxxxxxx 1 xx~~

~~xxxxxxx xx~~

~~xxxxxxxx xx y xx xxxxxxxx xx xx + xxxxxxx = + xxxxxxx x xx xx xx x x xx xx x x xx xx x xx xx x 2 3 2 3 x2 xx 3~~

~~xx xx x xx x x xx x xx xx x x xx x x x1 x4 1 4 1 4 xx x x xx y xx xx xx + xx + xx + + y xxy xx xx~~

~~x x xx xx xx xx x x xxx x xx xxx x23xx xx x 2323~~

~~x xx xx xx x 1 y 4~~

~~xx xx xx x1 xx xxx4 xx~~

~~xx + x xx xx y + +~~

~~xx xx x x xx x xx 3 2 x xx xx xxx x2xx 3~~

~~x1 xx xxx4~~

~~xx xx y + x xx xx~~

~~xx x x x2 3 (4) Fig. 5.3.4 - The graphical expansion up to the second order of Gc (x1,x2,x3,x4).~~

There are four ways to attach the first vertex to the first propagator and four for attaching the second vertex to the second propagator. Three ways for attaching the second leg of the first vertex to the second vertex and two ways of attaching the the third leg of the first vertex to the second one. Therefore we get 1 1 · 4 · 4 · 3 · 2= (5.56) 4! 6

134 The four point function can be evaluated in analogous way and the result is given in Fig. 5.3.4. Of course we can get it by functional differentiation of W [J]. The first order contribution is (4) 4 Gc (x1,x2,x3,x4)=−iλ d x∆F (x1 −x)∆F (x2 −x)∆F (x3 −x)∆F (x4 −x) (5.57) where the numerical coefficient is one, since there are 4 · 3 · 2 of attaching the vertex to the four external lines.

~~5.4 The Feynman’s rules in momentum space~~

The rules for the perturbative expansion of the Green’s functions in space-time that we got in the previous Section can be easily extended to the momentum space. We will consider again the two point function given in eq. (5.54) where we will take (2)(i) separately the various contributions in λ that we will denote by Gc ,wherei (2) stands for the power of λ. let us also deﬁne the Fourier transform of Gc (x ,x )as 1 2 (2) 4 4 (2) ip1x1 + ip2x2 Gc (p1,p2)= d x1d x2Gc (x1,x2)e (5.58)

Then we get i G(2)(0)(p ,p )=(2π)4δ4(p + p ) (5.59) c 1 2 1 2 2 − 2 p1 m + i and (2)(1) Gc (p1,p2)= λ d4p − − i = −i d4x d4x d4xeip1x1 + ip2x2 e ip(x1 x) 1 2 4 2 − 2 2 (2π) p m + i d4q i d4k − − i · e ik(x x2) 4 2 − 2 4 2 − 2 (2π) q m + i (2π) k m + i λ d4p d4q d4k = −i (2π)4δ4(p − k)(2π)4δ4(p − p)(2π)4δ4(p + k) 2 (2π)4 (2π)4 (2π)4 1 2 i i i · (5.60) p2 − m2 + i q2 − m2 + i k2 − m2 + i The product of the δ4 functions may be rewritten in such a way to pull out the overall four-momentum conservation 4 4 4 4 4 4 (2π) δ (p1 + p2)(2π) δ (p1 − p)(2π) δ (p − k) (5.61) By integrating over p and k we get λ i 2 d4q i G(2)(1)(p ,p )=−i (2π)4δ4(p +p ) (5.62) c 1 2 1 2 2 − 2 4 2 − 2 2 p1 m + i (2π) q m + i Also in momentum space we can make use of a diagrammatic expansion with the following rules:

~~135 • For each propagator draw a line with associated momentum p (see Fig. 5.4.1). . : ______i p 2 _ m2 + iε Fig. 5.4.1 - The propagator in momentum space.~~

• For each factor (−iλ/4!) draw a vertex with the convention that the momentum ﬂux is zero (see Fig. 5.4.2). . p p 1 4 λ : _ i __ , p + p + p + p = 0 4! 1 2 3 4 p p 23 Fig. 5.4.2 - The vertex in momentum space.

(n) • To get Gc draw all the topological inequivalent diagrams after having ﬁxed the external legs. The number of ways of drawing a given diagram is its topological weight. The contribution of such a diagram is multiplied by its topological weight.

• After the requirement of the conservation of the four-momentum at each vertex we need to integrate over all the independent internal four-momenta with weight d4q (5.63) (2π)4 A more systematic way is to associate a factor

4 iλ 4 4 − (2π) δ ( pi) (5.64) 4! i=1 to each vertex and integrate over all the momenta of the internal lines. This gives automatically a factor n 4 4 (2π) δ ( pi), (n = numero linee esterne) (5.65) i=1 corresponding to the conservation of the total four-momentum of the diagram.

~~By using these rules we can easily evaluate the 2nd order contribution of Fig. 5.4.3.~~

~~136 .~~

~~q 1 q p 2 p 1 2 q 3~~

~~(2) Fig. 5.4.3 - One of the second order contributions to Gc .~~

We get 2 3 4 × × × × −iλ i d qi i 4 4 3 2 2 2 4 2 2 4! p − m + i (2π) qi − m + i 2 i=1 i ·(2π)4δ4(p − q − q − q )(2π)4δ4(q + q + q − p ) 1 1 2 3 1 2 3 2 2 − 2 p1 m + i λ2 1 = (2π)4δ4(p + p ) 6 (p2 − m2 + i)2 1 2 1 3 4 d qi i · 4 4 − − − 4 2 2 (2π) δ (p1 q1 q2 q3) (5.66) (2π) qi − m + i i=1

~~5.5 Power counting in λϕ4~~

We start by going to the euclidean formulation. Let us denote the time by tR.The relation with the euclidean time is tE = itR. Recalling from the Section 4.9 that for the harmonic oscillator of unit mass one has (we have deﬁned xR =(tR,x), xE =(tE,x)) 3 3 d p 2 ip · x d p 2 ip · x i∆F (xR)= ∆(tR,ω )e = DE(tE,ω )e (2π)3 k (2π)3 k 4 −ipE xE d pE e = 4 2 2 (5.67) (2π) pE + m where pE =(EE,p), and we have used eq. (4.237) for unit mass and with ν = EE. From this we get the following modiﬁcations to Feynman’s rules in the euclidean version • Associate to each line a propagator 1 (5.68) p2 + m2

~~137 • To each vertex associate a factor λ − (5.69) 4! This rule follows because the weight in the euclidean path-integral is given by − e SE instead of eiS.~~

The other rules of the previous Section are unchanged. The reason to use the euclidean version is because in this way the divergences of the integrals become manifest by simply going in polar coordinates. We will now in the position to make a general analysis of the divergences. For simplicity we will consider only scalar particles. First of all consider the number of independent momenta in a given diagram, or the number of ”loops”, L.This is given, by deﬁnition, by the number of momenta upon which we are performing the integration, since they are not ﬁxed by the four-momentum conservation at the vertices. Since one of this conservation gives rise to the overall four-momentum conservation we see that the number of loops is given by

L = I − (V − 1) = I − V + 1 (5.70) where I is the number of internal lines and V the number of vertices. As in Section 2.3 we will be interested in evaluating the superﬁcial degree of divergence of a diagram. To this end notice that • L d The L independent integrations give a factor k=1 d pk where d is the number of space-time dimensions.

~~• Each internal line gives rise to a propagator containing two inverse powers in momenta I 1 2 2 (5.71) pi + m i=1 Given that, the superﬁcial degree of divergence is deﬁned as~~

~~D = dL − 2I (5.72)~~

Notice that D has to do with a common scaling of all the integration variables, therefore it may well arise the situation where D<0 (corresponding to a convergent situation), but there are divergent sub-integrations. Therefore we will need a more accurate analysis that we will do at the end of this Section. For the moment being let us stay with this deﬁnition. We will try now to get a relation between D and the number of external legs in a diagram. By putting VN equal to the total number of vertices with N lines we have (see Section 2.3)

~~NVN = E +2I (5.73)~~

138 with E the number of external legs and I the number of internal ones (recall we are dealing with scalar particles only). Eliminating I between this relation and the previous one, we get 1 I = (NVN − E) (5.74) 2 from which d − 2 D = d(I −VN +1)−2I =(d−2)I −d(VN −1) = (NVN −E)−d(VN −1) (5.75) 2 This relation can be rewritten as 1 N − 2 D = d − (d − 2)E + d − N VN (5.76) 2 2 In particular, for d =4wehave

~~D =4− E +(N − 4)VN (5.77)~~

~~For the λϕ4 theory this relation becomes particularly simple since then D depends only on the number of external legs~~

~~D =4− E (5.78)~~

Therefore the only diagrams superﬁcially divergent, D ≥ 0, are the ones with E =2 and E =4. As noticed before, this does not prove that a diagram with E>4orD<0is convergent. Let us discuss more in detail this point. Let us consider the diagram of Fig. 5.5.1, and let us suppose that the diagrams 1 and 2 have degrees of superﬁcial divergence D1 and D2 respectively. Since they are connected by n lines, we have n − 1 independent momenta, and therefore .

~~1 2~~

~~DD 1 n 2~~

~~Fig. 5.5.1 - Two diagrams connected by n-lines.~~

~~D = D1 + D2 +4(n − 1) − 2n = D1 + D2 +2n − 4 (5.79)~~

~~Therefore we may have D<0withD1 and D2 positive, that is both divergent. However for n>1wehave D ≥ D1 + D2 (5.80)~~

139 and therefore in order to have D<0 with 1 or 2 being divergent diagrams, we need D1 or D2 to be negative enough to overcome the other. To exclude the case n =1 means to consider only the one-particle irreducible (1PI) Feynman’s diagrams. In Fig. 5.5.2 we give an example of a one-particle reducible diagram. .

~~xxxxxxxxx xxxxxxxxx~~

~~Fig. 5.5.2 - One-particle reducible Feynman’s diagram.~~

The reason of excluding these diagrams from our considerations is that the momentum of the internal line connecting the two diagrams is ﬁxed, and therefore it cannot give rise to divergences. Furthermore, the one-particle reducible diagrams can be easily reconstructed from the 1PI’s. After that we look for the diagrams having D<0 but yet divergent, in order to have a complete classiﬁcation of the divergences. The idea is simply to look for the reducibility of the diagram. We will now stay with the λϕ4 theory. Then, we can consider the 2-, 3-particle reducible diagrams. In fact, for the nature of the vertex it is impossible to have a connected 4-particle irreducible diagram. We begin with the 3-particle reducibility. In this case we look for diagrams of the type of Fig. 5.5.3. . 1

~~xxxxxxxxx xxxxxxxxx~~

~~N Fig. 5.5.3 - Three-particle reducible Feynman’s diagram.~~

~~In this case we have D1 =0andD2 =1− N, therefore~~

D = D1 + D2 +2=3− N (5.81) In order the superﬁcial degree of divergence is negative, we must have N>4. In this diagram we have a primitively divergent four point function and a diagram with D<0 which can be again analyzed in the same way. In analogous way we look for two-particle reducible diagrams with D<0 and we extract contributions of the forms given in Fig 5.5.4.

~~140 . 1~~

~~xxxxxxxxx xxxxxxxx~~

~~xxxxxxxx~~

~~xxxxxxxx xxxxxxxxx~~

~~N Fig. 5.5.4 - Two-particle reducible Feynman’s diagram.~~

~~In the ﬁrst case we have D1 =2andD2 =2− N, therefore~~

~~D = D1 + D2 =4− N (5.82)~~

~~Again this requires N>4. The second diagram has again D2 < 0 and we can iterate the procedure. In the second case we have D1 =0andD2 =2− N, therefore~~

D = D1 + D2 =2− N (5.83) requiring N>2. By proceeding in this way we see that truly convergent diagrams do not contain hidden two and four point divergent functions. What we have shown is that for the λϕ4 theory in 4 dimensions a 1PI Feynman diagram is convergent if D<0 and it cannot be split up into 3- or 2-particle reducible parts containing two-and four-point divergent functions. This turns out to be a general theorem due to Weinberg, valid in any field theory and it says that a Feynman diagram is convergent if its superficial degree of divergence and that of all its subgraphs are negative. In the case of the λφ4 theory the Weinberg’s theorem tells us that all the divergences rely in the two and four point functions. If it happens that all the divergent diagrams (in the primitive sense specified above) correspond to terms which are present in the original lagrangian, we say that the theory is renormalizable. In general we can look at this question by studying the equation for the superficial divergence that we got before 1 N − 2 D = d − (d − 2)E + d − N VN (5.84) 2 2

~~141 For d = 4 we recall D =4− E +(N − 4)VN (5.85)~~

We see that D increases with VN for N>4. Therefore we have inﬁnite divergent diagrams and the theory is not renormalizable. We can have a renormalizable scalar theory only for N ≤ 4, meaning that the dimension N of the vertex operator, ϕN , must be less or equal to 4. An interesting case is the one of the theory λϕ3 in 6 dimensions. For d =6weget

D =6− 2E +(2N − 6)VN (5.86) and for N =3 D =6− 2E (5.87) and the only divergent functions are the one- two- and three-point ones. In d =6 we have dim[ϕ] = 2 and again in order to have renormalizability we must have that the dimensions of the vertex operator must be less or equal to 6.

~~5.6 Regularization in λϕ4~~

In this Section we will evaluate the one-loop contributions to the two- and four- point Green’s functions by using the dimensional regularization. To this end, let us consider the action in dimensions d =2ω 2ω 1 µ 1 2 2 λ 4 S ω = d x ∂µϕ∂ ϕ + m ϕ + ϕ (5.88) 2 2 2 4!

Inourunitsystemwehave(/h = c = 1), and therefore the action S2ω should be dimensionless. As discussed in Section 2.3 this allows to evaluate the dimensions of the fields, and we recall that for the scalar field we have d dim[ϕ]= − 1=ω − 1 (5.89) 2 From this it follows at once that dim[m]=1,dim[λ]=4− 2ω (5.90) As we did in QED for the electric charge, here too it is convenient to introduce a dimensionful parameter µ in order to be able to define a dimensionless coupling

~~2 ω−2 λnew = λold(µ ) (5.91)~~

~~Then the action will be (λ = λnew) 2ω 1 µ 1 2 2 λ 2 2−ω 4 S ω = d x ∂µϕ∂ ϕ + m ϕ + (µ ) ϕ (5.92) 2 2 2 4! The Feynman’s rules are slightly modiﬁed~~

142 • The scalar product among two four-vectors becomes a sum over 2ω components. • In the loop integrals we will have a factor d2ωp (5.93) (2π)2ω and the δ function in 2ω-dimensions will be deﬁned by 2ω (2ω) d pδ ( pi) = 1 (5.94) i

• The coupling λ goes into λ(µ2)2−ω. Also, we will do all the calculation in the euclidean version. Remember that in Section 2.5 we gave the relevant integrals both in the euclidean and in the minkowskian version. We start our calculation from the ﬁrst order contribution to the two-point function corresponding to the diagram (tadpole) of Fig. 5.6.1. This is given by 1 1 π 2ωδ(2ω) p p T (2 ) ( 1 + 2) 2 2 2 2 2 (5.95) p1 + m p1 + m . p

~~p p 1 2~~

Fig. 5.6.1 - The one-loop contribution to the two-point function in the λϕ4 theory. with 1 d2ωp 1 λ πω 1 T = (−λ)(µ2)2−ω = − (µ2)2−ω Γ(1 − ω) 2 2 (2π)2ω p2 + m2 2 (2π)2ω (m2)1−ω −ω λ m2 4πµ2 2 = − Γ(1 − ω) (5.96) 2 (4π)2 m2 For ω → 2weget λ m2 4πµ2 1 T ≈− 1+(2− ω)log + ··· × (−1) + ψ(2) + ··· 2 2 2 − 2 (4π) m 2 ω λ 1 4πµ2 = m2 + ψ(2) + log + ··· (5.97) 32π2 2 − ω m2

~~143 . p p 1 p - k 4~~

~~p k p 2 3 Fig. 5.6.2 - The one-loop contribution to the four-point function in the λϕ4 theory.~~

This expression has a simple pole at ω = 2 and a ﬁnite part depending explicitly on µ2. Let us consider now the one-loop contribution to the four-point function. The corresponding diagram is given in Fig. 5.6.2, with an amplitude

~~4 2ω (2ω) 1 (2π) δ (p1 + p2 + p3 + p4) 2 2 T4 (5.98) pk + m k=1 with 1 d2ωk 1 1 T = (−λ)2(µ2)4−2ω ,p= p + p (5.99) 4 2 (2π)2ω k2 + m2 (k − p)2 + m2 1 2~~

We reduce the two denominator to a single one, through the equation (see Section 2.6) 1 1 1 1 = dx k2 + m2 (k − p)2 + m2 [x(k2 + m2)+(1− x)((k − p)2 + m2)]2 0 1 1 = dx [x(k2 + m2)+(1− x)(k2 + m2 − 2k · p + p2)]2 0 1 1 = dx (5.100) 2 2 2 2 0 [(k + m )+(1− x)(p − 2k · p)] the denominator can be written as

~~(k2 + m2)+(1− x)(p2 − 2k · p)=[k − (1 − x)p]2 + m2 + p2(1 − x) − p2(1 − x)2 =[k − (1 − x)p]2 + m2 + p2x(1 − x) (5.101)~~

~~Since our integral is a convergent one, it is possible to translate k into k − p(1 − x) obtaining 2 1 2ω λ 2 4−2ω d k 1 T4 = (µ ) dx (5.102) 2ω 2 2 2 2 2 0 (2π) [k + m + p x(1 − x)]~~

144 By performing the momentum integration we get 2 1 λ 2 4−2ω dx 1 T4 = (µ ) Γ(2 − ω) (5.103) ω 2 2 2−ω 2 0 (4π) [m + p x(1 − x)] In the limit ω → 2 λ2 1 dx 1 T ≈ (µ2)2−ω 1+(2− ω)logµ2 + ψ(1) 4 2 − 2 0 (4π) 2 ω m2 + p2x(1 − x) × 1 − (2 − ω)log 4π λ2 1 1 m2 + p2x(1 − x) ≈ (µ2)2−ω + ψ(1) − dx log (5.104) 2 − 2 32π 2 ω 0 4πµ Also the x integration can be done by using the formula √ 1 4 √ 1+a +1 dx log 1+ x(1 − x) = −2+ 1+a log √ ,a>0 (5.105) 0 a 1+a − 1 The ﬁnal result is λ2 1 4πµ2 T ≈ (µ2)2−ω + ψ(1) + 2 + log 4 32π2 2 − ω m2 4m2 4m2 1+ p2 +1 − 1+ log + ··· (5.106) p2 4m2 − 1+ p2 1

~~5.7 Renormalization in the theory λϕ4~~

~~In order to renormalize the theory λϕ4 we proceed as in QED. That is we split the original lagrangian, expressed in terms of bare couplings and bare ﬁelds in a part~~

~~1 µ 1 2 2 λ 4−2ω 4 Lp = ∂µϕ∂ ϕ + m ϕ + µ ϕ (5.107) 2 2 4! depending on the renormalized parameters m and λ and ﬁeld ϕ,andinthecounter- terms contribution~~

~~1 µ 1 2 2 δλ 4−2ω 4 Lct = δZ∂µϕ∂ ϕ + δm ϕ + µ ϕ (5.108) 2 2 4! The sum of these two expressions~~

~~1 µ 1 2 2 2 (λ + δλ) 4−2ω 4 L = (1 + δZ)∂µϕ∂ ϕ + (m + δm )ϕ + µ ϕ (5.109) 2 2 4!~~

~~145 gives back the original lagrangian, if we redeﬁne couplings and ﬁelds as follows~~

~~1/2 1/2 ϕB =(1+δZ) ϕ ≡ Zϕ ϕ (5.110)~~

~~m2 + δm2 m2 = =(m2 + δm2)Z−1 (5.111) B 1+δZ ϕ~~

4−2ω λ + δλ 4−2ω −2 λB = µ = µ (λ + δλ)Z (5.112) (1 + δZ)2 ϕ In fact 1 µ 1 2 2 λ0 4 L = ∂µϕB∂ ϕB + m ϕ + ϕ (5.113) 2 2 B B 4! B The counter-terms are then evaluated by the requirement of removing the divergences arising in the limit ω → 2, except for a ﬁnite part which is ﬁxed by the renormalization conditions. From the results of the previous Section, by summing up .

~~xxxxxxxx~~

~~xx xxxxxxxx xx xx xx xx xx xx~~

~~xxxxxxxx = + +~~

~~xxxxxxxx~~

xx xx xx xx xxxxxx + xx xx xx xx + xxxxxx Fig. 5.7.1 - The contributions to the teo-point connected Green’s function. all the 1PI contributions we get the following expression for the two-point connected Green’s function (see Fig. 5.7.1) 1 1 1 (2) 2ω 2ω ··· Gc (p1,p2)=(2π) δ (p1 + p2) 2 2 + 2 2 T2 2 2 + p1 + m p1 + m p1 + m 1 ≈ (2π)2ωδ2ω(p + p ) (5.114) 1 2 2 2 − p1 + m T2 Therefore, the eﬀect of the ﬁrst order contribution is to produce a mass correction 2 determined by T2. The divergent part can be taken in care by the counterterm δm which produces an additional contribution to the one-loop diagram, represented in Fig. 5.7.2. We choose the counterterm as 1 λ 2 m2 δm2ϕ2 = m2 + F ω, ϕ2 (5.115) 2 64π2 1 µ2 with =4−2ω and F1 an arbitrary dimensionless function Adding the contribution of the counterterm to the previous result we get 1 1 1 G(2)(p ,p )=(2π)2ωδ2ω(p + p ) + (−δm2) c 1 2 1 2 2 2 − 2 2 2 2 p1 + m T2 p1 + m p1 + m (5.116)

~~146 .~~

~~_ 2 xx x = δ m~~

Fig. 5.7.2 - The Feynman’rule for the counterterm δm2. at the order λ we have 1 G(2)(p ,p )=(2π)2ωδ2ω(p + p ) (5.117) c 1 2 1 2 2 2 2 − p1 + m + δm T2 Of course the same result could have been obtained by noticing that the counterterm 2 2 2 2 modiﬁes the propagator by shifting m into m +δm . Now the combination δm −T2 is ﬁnite and given by λ m2 δm2 − T = m2 F − ψ(2) + log (5.118) 2 32π2 1 4πµ2

~~The ﬁnite result at the order λ is then~~

~~(2) 2ω 2ω 1 Gc (p1,p2)=(2π) δ (p1 + p2) 2 (5.119) 2 2 λ − m p1 + m 1+ 32π2 F1 ψ(2) + log 4πµ2~~

~~This expression has a pole in Minkowski space, and we can deﬁne the renormalization condition by requiring that the inverse propagator at the physical mass is~~

~~2 2 p + mphys (5.120)~~

This choice determines uniquely the F1 term. However we will discuss more in detail the renormalization conditions in the next Chapter when we will speak about the renormalization group. Notice that at one loop there is no wave function renormalization, but this comes about at two loops, as we shall see later. Consider now the four-point connected Green’s function. Its diagrammatic expansion at one loop is given in Fig. 5.7.3 (again we have included only 1PI diagrams).

~~From the results of the previous Section and introducing the Mandelstam invariants 2 2 2 s =(p1 + p2) ,t=(p1 + p3) ,u=(p1 + p4) (5.121) we get~~

~~4 (4) 2ω (2ω) 1 Gc (p1,p2,p3,p4)=(2π) δ (p1 + p2 + p3 + p4) 2 2 pk + m k=1 3λ 2 m2 1 × (−µ4−2ω)λ 1 − + ψ(1) + 2 − log − A(s, t, u) (5.122) 32π2 4πµ2 3~~

~~147 .~~

~~xx xx~~

~~xx xx xx xx~~

~~xx xxxxxxx xx xx~~

~~xxxxxxx~~

~~xxxxxxxxx x~~

~~xxxxxxxxx x xx x xxxxxxx = + + xxxxxxx~~

~~xx xxxxxxx~~

~~xx xx xx x~~

~~xx x~~

~~xx x xx~~

~~xx xx + + xx~~

~~xx xx~~

~~xx xx xx~~

xx Fig. 5.7.3 - The one-loop expansion for the four-point Green’s function. with 4m2 4m2 1+ z +1 A(s, t, u)= 1+ log (5.123) m2 z s,t,u z 4 − = 1+ z 1 The divergent part can be disposed oﬀ by the counterterm δλ 1 3λ2 2 m2 µ4−2ωϕ4 = µ4−2ω + G ω, ϕ4 (5.124) 4! 4! 32π2 1 µ2 with G1 an arbitrary dimensionless function. This counterrtem produces an additional contribution with a Feynman rule represented in Fig. 5.7.4. . _ 2 ω = _ ( µ 2 ) δλ

~~Fig. 5.7.4 - The Feynman’s rule for the couterterm δλ.~~

~~By adding the counterterm we get~~

~~4 1 G(4) p ,p ,p ,p π 2ωδ(2ω) p p p p −µ2−4ω λ c ( 1 2 3 4)=(2 ) ( 1 + 2 + 3 + 4) 2 2 ( ) pk + m k=1 3λ m2 1 × 1 − −G + ψ(1) + 2 − log − A(s, t, u) (5.125) 32π2 1 4πµ2 3~~

~~A possible renormalization condition is to ﬁx the scattering amplitude at some value 2 of the invariants (for instance, s =4mphys, t = u = 0) to be equal to the lowest order value~~

~~4 1 (4) 2ω (2ω) × − 2−4ω Gc (p1,p2,p3,p4)=(2π) δ (p1 + p2 + p3 + p4) 2 2 ( µ )λ (5.126) pk + m k=1~~

148 In any case G1 is completely ﬁxed by this condition. Without entering into the calculation we want now show how the renormalization program works at two loops. We will examine only the two-point function. Its complete expansion up to two loops, including also the counterterm contributions is given in Fig. 5.7.5. By omitting the external propagators and the delta-function of .

~~xxxxxxxx~~

~~xx xxxxxxxx xx xx xx xx xx xx xx xxxxxxxx xx = xx xx + xx xx xx + xxxxxxxx~~

~~xxxxxxxx~~

~~xx xx xx xx xx xx xx xx x~~

~~+ + + xx xx x + a) b)~~

+ xx xx + xx x x c) d) Fig. 5.7.5 - The two-loop expansion for the two-point Green’s function. conservation of the four-momentum, the various two-loop contributions are (notice that a counterterm inside a loop gives a second order contribution as a pure two- loops diagram):

~~• Contribution from the diagram a) of Fig 5.7.5 λ2 1 6m2 6m2 3 m2 − p2 + + + ψ(1) − log (5.127) 6(16π2)2 2 2 2 4πµ2~~

~~• Contribution from the diagram b) of Fig 5.7.5 λ2 4 2 m2 − + ψ(2) + ψ(1) − 2log (5.128) 4(16π2)2 2 4πµ2~~

~~• Contribution from the diagram c) of Fig 5.7.5 3λ2 4 2 m2 m2 + ψ(2) + G − log (5.129) 4(16π2)2 2 1 4πµ2~~

~~• Contribution from the diagram d) of Fig 5.7.5 λ2 4 2 m2 m2 + ψ(1) + F − log (5.130) 4(16π2)2 2 1 4πµ2~~

~~149 Adding together these terms we ﬁnd~~

λ2 p2 2λ2 1 λ2 1 − + m2 + m2 [F +3G − 1] (5.131) 12(16π2)2 (16π2)2 2 2(16π2)2 1 1 wherewehaveusedψ(2) − ψ(1) = 1. This expression is still divergent but at the order λ2. Therefore we can make ﬁnite the diagram by modifying at the second order the δm2 counterterm 1 δm2 + δm2 ϕ2 (5.132) 2 1 2 2 with δm1 given in eq. (5.115) and λ2 4 1 m2 δm2 = m2 + (F +3G − 1) + F ω, (5.133) 2 2(16π2)2 2 1 1 2 µ2

~~2 and F2 a new dimensionless function. The further divergence in p can be eliminated through the wave function renormalization counterterm, that is~~

1 µ δZ∂µϕ∂ ϕ (5.134) 2 with the choice λ2 2 m2 δZ = − + H (ω, ) (5.135) 24(16π2)2 2 µ2 with H2 dimensionless. All the functions F1, F2, G1 H2 are finite in the limit ω → 2. Notice also that we are not introducing more and more freedom, because only particular combinations of these functions appear in the counterterms, and are precisely these combinations which are fixed by the renormalization conditions. Also, the function H2 is fixed by the renormalization condition on the inverse propagator, which are really two conditions: a) the position of the pole, b) the residuum at the pole. .

~~Fig. 5.7.6 - Insertion of counterterms in multi-loop diagrams.~~

The previous procedure can be easily extended to higher and higher orders in the expansion in the coupling constant. What makes the theory renormalizable is the fact that the new divergences that one encounters can all be disposed oﬀ by modifying the counterterms δm2 and λ. This depends from the fact that the divergent terms are constant or quadratic in the external momentum, and therefore 2 µ 4 they can be reproduced by operators as ϕ , ∂µϕ∂ ϕ and ϕ . The non trivial part of

150 the renormalization program is just the last one. In fact consider diagrams of the type represented in Fig. 5.7.6, where we have taken out all the possible external lines. These diagrams contain 1/ terms coming from the counterterms times log p2 terms from the loop integration. In the previous two-loop calculation we had no such terms but we got their strict relatives log(m2/4πµ2). If present the log p2 terms would be a real disaster since they are non-local in conﬁguration space and as such they cannot be absorbed by local operators. However if one makes all the required subtractions, that is one subtracts correctly also the diagrams of the type in Fig. 5.7.6, it is possible to show that the log p2 terms cancel out. For instance, 2 in the expression for δm2 there are not such terms. In fact they cancel in adding up the four two-loop contributions.

~~151 Chapter 6~~

~~The renormalization group~~

~~6.1 The renormalization group equations~~

For the following considerations it will be convenient to introduce the one-particle (n) irreducible truncated n-point functions Γ (p1,p2, ···,pn) deﬁned by removing the propagators on the external legs and omitting the delta-function expressing the conservation of the four-momentum n n 2ω (2ω) (n) ··· (n) ··· −1 (2π) δ ( pi)Γ (p1,p2, ,pn)=G1PI(p1,p2, ,pn) DF (pi) (6.1) i=1 i=1 where 2ω ipx DF (p)= d xe 0|T (ϕ(x)ϕ(0))|0 (6.2) is the exact euclidean propagator. We have two ways of evaluating these functions, we can start from the original lagrangian and evaluate them in terms of the bare 4 couplings constants (in the case of λϕ , λB and mB). On the other hand we can express the bare parameters in terms of the renormalized ones, and, at the same time, renormalizing the operators through the wave function renormalization (ϕB = 1/2 Zϕ ϕ), obtaining the relation

~~(n) −n/2 (n) ΓB (p1,p2, ···,pn; λB,mB,ω)=Zϕ Γ (p1,p2, ···,pn; λ, m, µ, ω) (6.3)~~

−n/2 n/2 The factor Zϕ originates from a factor Zϕ coming from the renormalization (n) −1 factors in Gc and from n factors Zϕ from the inverse propagators. Of course, in a renormalizable theory the Γ(n)’s are finite by construction. In the previous relation we must regard the renormalized parameters as function of the bare ones. In the 4 λϕ case, this means λ and m depending on λB and mB. However the right hand side of the previous equation does not depend on the choice of the parameter µ. This implies that also the functions Γ(n) do not depend on µ except for the field −n/2 rescaling factor Zϕ . These considerations can be condensed into a differential

152 equation obtained by requiring that the total derivative with respect to µ of the right hand side of eq. (6.3) vanishes ∂ ∂λ ∂ ∂m ∂ n ∂ log Zϕ (n) µ + µ + µ − µ Γ (p ,p , ···,pn; λ, m, µ, ω)=0 ∂µ ∂µ ∂λ ∂µ ∂m 2 ∂µ 1 2 (6.4) The content of this relation is that a change of the scale µ can be compensated by a convenient change of the couplings and of the field normalization. Notice that since (n) Γ is finite in the limit ω → 2, it follows that also the derivative of log Zϕ is finite. It is convenient to define the following dimensionless coefficients ∂λ m µ = β λ, ,ω ∂µ µ 1 ∂ log Zϕ m µ = γd λ, ,ω 2 ∂µ µ µ ∂m m = γm λ, ,ω (6.5) m ∂µ µ obtaining ∂ ∂ ∂ (n) µ + β + γmm − nγd Γ (p ,p , ···,pn; λ, m, µ, ω) = 0 (6.6) ∂µ ∂λ ∂m 1 2

We can try to eliminate from this equation the term ∂/∂µ. To this end let us consider the dimensions of Γ(n).Indimensionsd =2ω the delta function has dimensions −2ω and dim[ϕ]=ω − 1, therefore n (n) 2ω i(p1x1 + ···+ pnxn) Gc (p1, ···,pn)= d xie 0|T (ϕ(x1) ···ϕ(xn)|0 conn. i=1 (6.7) has dimensions n(ω − 1) − 2nω = −n(ω + 1). The propagator DF (p) has dimensions 2(ω − 1) − 2ω = −2, and we get dim Γ(n) − 2ω = −n(ω +1)+2n → dim Γ(n) = n + ω(2 − n) (6.8) and putting =4− 2ω dim Γ(n) =4− n + (n − 2) (6.9) 2 Therefore Γ(n) must be a homogeneous function of degree 4 − n + (n − 2)/2inthe dimensionful parameters pi, m e µ. Introducing a common scale s for the momenta we get (from the Euler theorem) ∂ ∂ ∂ (n) µ + s + m − (4 − n + (n − 2)) Γ (spi; λ, m, µ, ) = 0 (6.10) ∂µ ∂s ∂m 2

153 Since everything is ﬁnite for → 0 we can take the limit and eliminate ∂/∂µ between this equation and eq. (6.6) ∂ ∂ ∂ (n) −s + β +(γm − 1)m − nγd +4− n Γ (spi; λ, m, µ, ) = 0 (6.11) ∂s ∂λ ∂m

This equation tells us how the Γ(n) scale with the momenta, and allows us to evaluate them at momenta spi once we know them at momenta pi. To this end one needs to know the coefficients β, γm and γd. For this purpose one needs to specify the renormalization conditions since these coefficients depend on the arbitrary functions F1,G1, ···. In order to use the renormalization group equations we need to know more about the structure of the functions β, γd and γm. We have seen that the counterterms are singular expressions in ,therefore it will be possible to express them as a Laurent series in the renormalized parameters ∞ ak λB = µ a + (6.12) 0 k k=1 ∞ 2 2 bk mB = m b0 + k (6.13) k =1 ∞ ck Zϕ = c + (6.14) 0 k k=1 with the coefficients functions of the dimensionless parameters λ and m/µ.Notice that a0, b0 and c0 may depend on but only in an analytical way. Recalling from the previous Section the expressions for the counterterms at the second order in λ 3λ2 2 δλ = + G (6.15) 32π2 1 λ 2 λ2 4 1 δm2 = m2 + F + m2 + (F +3G − 1) + F (6.16) 32π2 1 2(16π2)2 2 1 1 2 λ2 2 δZ = − + H (6.17) 24(16π2)2 2 we get for the bare parameters 2 3λ 2 3 λB = µ λ + + G + O(λ ) (6.18) 32π2 1

154 λ 2 m2 m2 F B = 1+ 2 + 1 32π λ2 4 1 + + (F +3G − 1) + F 2(16π2)2 2 1 1 2 λ2 2 + + H + O(λ3) (6.19) 24(16π2)2 2 2 λ 2 3 Zϕ =1− + H + O(λ ) (6.20) 24(16π2)2 2 Therefore the coeﬃcients are given by

3λ2 a = λ + G + O(λ3) (6.21) 0 32π2 1 3λ2 a = + O(λ3) (6.22) 1 16π2 λ λ2 1 b =1+ F + (F + H )+O(λ3) (6.23) 0 32π2 1 2(16π2)2 2 12 2 λ λ2 λ2 b = + (F +3G − 1) + + O(λ3) (6.24) 1 16π2 2(16π2)2 1 1 12(16π2)2 2λ2 b = + O(λ3) (6.25) 2 (16π2)2 λ2 c =1− H + O(λ3) (6.26) 0 24(16π2)2 2 λ2 c = − + O(λ3) (6.27) 1 12(16π2)2 Notice that the dependence on m/µ comes only from the arbitrary functions defining the finite parts of the counterterms. This is easily understood since the divergent terms originate from the large momentum dependence where the masses can be neglected. This is an essential feature, since it is very difficult to solve the group renormalization equations when the coefficients β, γm e γd depend on both variables λ and m/µ. On the other hand since these coefficients depend on the renormalization conditions, it is possible to define a scheme where they do not depend on m/µ. Another option would be to try to solve the equations in the large momentum limit where the mass dependence can be neglected. Callan and Szymanzik have considered a different version of the renormalization group equations. They have considered the dependence of Γ(n) on the physical mass. In that case one can show that β and γd depend only on λ. Furthermore in their equation the terms γm and µ/∂µ do not appear. In their place there is a inhomogeneous term in the large momentum limit.

~~155 6.2 Renormalization conditions~~

Let us consider again the renormalization conditions. The renormalization procedure introduces in the theory a certain amount of arbitrariness in terms of the scale µ and of the arbitrary functions F1,F2 ···. Recall that the lagrangian has been decomposed as L = Lp + Lct (6.28) therefore, the finite part of Lct can be absorbed into a redefinition of the parameters appearing in L since both have the same operatorial structure. This is another way of saying that the finite terms in Lct arefixedonceweassignλ and m. Inturnthese are fixed by relating them to some observable quantity as a scattering cross-section. Furthermore we have to fix a scale µ at which we define the parameters λ and m. The renormalization group equations then tell us how to change the couplings when we change µ, in order to leave invariant the physical quantities. As we have said the procedure for fixing m and λ is largely arbitrary and we will exhibit various possibilities. To do that it is convenient to write explicitly the expression for the two- and four-point truncated amplitudes λ2 Γ(2)(p)= 1 − H p2 24(16π2)2 2 λ m2 + m2 1+ F − ψ(2) + log 32π2 1 4πµ2 λ2 + F + ··· (6.29) 2(16π2)2 2 3λ m2 1 Γ(4)(p ,p ,p ,p )=−µλ 1 − ψ(1) + 2 − log − A(s, t, u) − G 1 2 3 4 32π2 4πµ2 3 1 (6.30) In the two point function we have omitted the finite terms coming from the second order contributions for which we have given only the divergent part. let us now discuss various possibilities of renormalization conditions (sometimes one speaks of subtraction conditions) • A first possibility is to renormalize at zero momenta, that is requiring (2) 2 2 Γ (p) = p + mA (6.31) p2∼0 (4) Γ (p1,p2,p3,p4) = −µ λA (6.32) pi=0 2 where mA and λA are quantities to be fixed by comparing some theoretical cross-section with the experimental data. Then, comparing with eqs. (6.29) and (6.30) and using A(s, t, u)=6forpi =0,weget m2 m2 HA =0,FA = ψ(2) − log A ,GA = ψ(1) − log A (6.33) 2 1 4πµ2 1 4πµ2

156 Another possibility would be to ﬁx the momenta at some physical value in the Minkowski space. For instance we could require Γ(−p2 =¯m2)=0where¯m is the physical mass and then ﬁx Γ(4) with all the momenta on shell. • To choose the subtraction point at zero momenta is generally dangerous for massless particles (in this way one would introduce spurious infrared divergences). As a consequence a popular choice is to subtract at an arbitrary value of the momenta. For instance one requires

~~(2) 2 2 2 2 Γ (p)=p + mB,p= M (6.34)~~

(4) 2 Γ (p1,p2,p3,p4)=−µ λB,s= t = u = M (6.35) Notice that the point chosen for the four-point function is a non-physical one. In this case we get m2 m2 1 HB =0,FB = ψ(2)−log B ,GB = ψ(1)+2−log B − A(M 2,M2,M2) 2 1 4πµ2 1 4πµ2 3 (6.36) This way of renormalize has however the problem of leaving an explicit dependence on the mass in the coefficients of the renormalization group equations. • The last way of renormalizing we will consider is due to Weinberg and ’t Hooft and the idea is simply to put equal to zero all the arbitrary functions F1,F2, ··· at each order in the expansion. In this way the coefficients ai, bi and ci appearing in the Laurent series defining the bare couplings turn out to be mass independent. The Weinberg and ’t Hooft prescription allows to evaluate the renormalization group equations coefficients in a very simple way. For instance, let us start with the bare coupling ∞ ak(λ) λB = µ λ + (6.37) k k=1 Differentiating with respect to µ and recalling that λB does not depend on this variable, we get ∞ ∞ ak(λ) ∂λ a (λ) 0= λ + + µ 1+ k (6.38) k ∂µ k k=1 k=1 where dak(λ) a (λ)= (6.39) k dλ Since β = µ∂λ/∂µ is analytic for → 0, for small we can write β = A + B, obtaining ∞ ∞ ak(λ) a (λ) 0= λ + +(A + B) 1+ k (6.40) k k k=1 k=1

~~157 and identifying the coeﬃcients of the various powers in ~~

~~B + λ = 0 (6.41)~~

a1 + A + Ba1 = 0 (6.42) ak+1 + Aak + Bak+1 = 0 (6.43) from which B = −λ (6.44) d A = − 1 − λ a (6.45) dλ 1 We then get d β(λ)=− 1 − λ a − λ (6.46) dλ 1 and for → 0 d β(λ)=− 1 − λ a (6.47) dλ 1 and d dak d 1 − λ ak = 1 − λ a (6.48) dλ +1 dλ dλ 1

From the expression of the previous Section for a1 at the second order in λ,weget ∂λ 3λ2 µ = β(λ)= (6.49) ∂µ 16π2 This equation can be integrated in order to know how we have to change λ when we change the value of the scale µ. The function λ(µ) is known as the running coupling constant. Its knowledge, together with the knowledge of m(µ) allows us to integrate easily the renormalization group equations. In fact, let us deﬁne

~~t =logs (6.50) and let us consider the following partial diﬀerential equation (we shall see that the renormalization group equations can be brought back to this form)~~

∂ ∂ F (xi,t)=βi(xi) F (xi,t) (6.51) ∂t ∂xi The general solution to this equation is obtained by using the method of the ”char- acteristics”. That is one considers the integral curves (the characteristic curves) deﬁned by the ordinary diﬀerential equations

~~dxi(t) = βi(xi(t)) (6.52) dt~~

~~158 with given boundary conditions xi(0) =x ¯i. Then, the general solution is~~

F (xi,t)=ψ(xi(t)) (6.53) where ψ(¯xi)=F (¯xi, 0) (6.54) is an arbitrary function to be determined by the condition at t =0requiredforF . In fact we can check that ψ satisﬁes the diﬀerential equation

∂ dxi(t) ∂ψ ∂ψ ψ(xi(t)) = = βi (6.55) ∂t dt ∂xi ∂xi In our case the equation contains a non-derivative term ∂ ∂ G(xi,t)=βi(xi) G(xi,t)+γ(xi)G(xi,t) (6.56) ∂t ∂xi with γ(xi) a given function. By putting

~~t dt γ(xi(t )) G(xi,t)=e 0 F (xi,t) (6.57) we see that F (xi,t) satisﬁes eq. (6.51) and therefore the general solution will be~~

~~t dt γ(xi(t )) G(xi,t)=e 0 ψ(xi(t)) (6.58)~~

~~(n) with xi(t) satisfying eq. (6.52). For the Γ we get (t =logs)~~

s − ds n γd(λ(s )) (n) (n) 4−n Γ (spi; m, λ, µ)=Γ (pi; m(s),λ(s),µ)s e 1 s (6.59) with ∂λ(s) s = β(λ(s)) (6.60) ∂s ∂m(s) s = m(s)(γm(λ(s)) − 1) (6.61) ∂s and λ(s =1)=λ, m(s =1)=m. This result tells us that when we perform a re-scaling on the momenta, the amplitudes Γ(n) do not scale only with the trivial factor s4−n due to their dimensions, but that they undergo also a non trivial scaling, due to γd = 0, which is necessary to compensate the variations of λ and m with the scale.

~~159 6.3 Application to QED~~

In this Section we will apply the previous results to QED. In particular, we want to check that the running coupling obtained via the renormalization group equations is the same that we have defined in Section 2.2. First of all let us start with the QED lagrangian in d =2ω dimensions (neglecting the gauge fixing terms) 2ω 1 µν S ω = d x ψ¯(i∂ˆ − m)ψ − eψ¯Aψˆ − Fµν F (6.62) 2 4 Since 1 dim[ψ¯Aψˆ ]=ω − 1+2 ω − =3ω − 2 (6.63) 2 it follows dim[e]=2− ω (6.64) Therefore we define ω−2 −/2 enew = eold(µ) = eoldµ (6.65)

As before we write the lagrangian as Lp + Lct, with (we ignore here the gauge ﬁxing term) /2 1 µν Lp = ψ¯(i∂ˆ − m)ψ − eµ ψ¯Aψˆ − Fµν F (6.66) 4 and /2 δZ3 µν Lct = iδZ ψ¯∂ψˆ − δmψψ¯ − eδZ µ ψ¯Aψˆ − Fµν F (6.67) 2 1 4 giving

/2 1+δZ3 µν L = i(1 + δZ )ψ¯∂ψˆ − (m + δm)ψψ¯ − e(1 + δZ )µ ψ¯Aψˆ − Fµν F (6.68) 2 1 4 After re-scaling of the ﬁelds one gets back the original lagrangian via, in particular, the identiﬁcation

/2 (1 + δZ1) /2 eB = µ e = µ e(1 − δZ3/2) (6.69) (1 + δZ2)(1 + δZ3/2) wherewehaveusedδZ1 = δZ2. From Section 2.7 we recall that e2 δZ = C = − (6.70) 3 6π2 and therefore 2 /2 e eB = µ e 1+ (6.71) 12π2 We proceed as in the previous Section by deﬁning /2 ak eB = µ e + k (6.72) k

160 with e3 a = (6.73) 1 12π2 Requiring that eB is µ-independent, and in the limit → 0weget ∂e 1 d β(e)=µ = − 1 − e a (6.74) ∂µ 2 de 1 from which e3 β(e)= (6.75) 12π2 The diﬀerential equation deﬁning the running coupling constant is ∂e(µ) e3 µ = β(e)= (6.76) ∂µ 12π2 Integrating this equation we obtain 1 1 = − log µ2 + K (6.77) e2(µ2) 12π2 where K is an integration constant which can be eliminated by evaluating e at another momentum scale Q2.Weget 1 1 1 Q2 − = − log (6.78) α(Q2) α(µ2) 3π µ2 or α(µ2) α(Q2)= (6.79) − α(µ2) Q2 1 3π log µ2 which agrees with the result obtained in Section 2.2.

~~6.4 Properties of the renormalization group equations~~

Integrating eq. (6.49) dλ 3λ2 µ = + O(λ3) (6.80) dµ 16π2 we get 1 3 µ 1 − = log − (6.81) λ(µ) 16π2 µs λs where µs is a reference scale and λs = λ(µs). Therefore 1 λ(µ)=λs (6.82) 3λs µ 1 − 2 log 16π µs

161 From this expression we see that starting from µ = µs, λ increases with µ. Suppose that we start with a small λ in such a way that the perturbative expansion is sensible. However increasing µ we will need to add more and more terms to the expansion due to the increasing of λ. Therefore the λϕ4 theory allows a perturbative expansion only at small mass scales, or equivalently at large distances. In this theory we can give a definite sense to the asymptotic states. The increasing of λ is related to the sign of the β(λ) function. For the λϕ4 theory and for QED the β function is positive, however if it would be negative the theory would become perturbative at large mass scales or at small distances. In this case one could solve the renormalization group equations at large momenta by using the perturbative expansion at a lower order to evaluate the β and the other coefficients. However in this case the coupling is increasing at large distances and this would be a problem for our approach based on th in- and out- states. As we shall see this is the situation in QCD. here the fields that are useful to describe the dynamics at short distances (the quark fields) cannot describe the asymptotic states, which can be rather described by bound states of quarks as mesons and baryons. Going back to the eq. (6.82) we see that, assuming the validity of the expression for the β function up to large momenta, the coupling constant becomes infinite at the value of µ given by 16π2

µ = µse 3λs (6.83) which is a very big scale if λs is small. The pole in eq. (6.82) is called the Landau pole (see Section 2.2 in the case of QED). Of course there are no motivations why eq. (6.82) should be valid for large values of λ. Let us now discuss several possible scenarios starting from a theory where β =0 for λ =0.

• If β(λ) > 0 for any λ, than the running coupling is always increasing, and it will become inﬁnite at some value of µ. If this happens for a ﬁnite µ we say that we have a Landau pole at that scale.

~~• Suppose that β(λ) is positive at small λ and that it becomes negative going through a zero at λ = λF (see Fig. 6.4.1)).~~

The point λF is called a fixed point since if we start with the initial condition λ = λF than λ remains fixed at that value. To study the behaviour of λ around the fixed point, let us expand the β(λ) around its zero

~~β(λ) ≈ (λ − λF )β (λF ) (6.84)~~

~~It follows dλ µ =(λ − λF )β (λF )+··· (6.85) dµ~~

~~162 . β(λ)~~

~~λ λ λ S S S λ λ F~~

~~Fig. 6.4.1 - The behaviour of β(λ) giving rise to an ultraviolet ﬁxed point.~~

~~from which dλ dµ = β (λF ) (6.86) λ − λF µ Integrating this equation we get~~

~~β (λF ) λ − λF µ = (6.87) λs − λF µs~~

Notice that the sign of β (λF ) plays here a very important role. In the present case the sign is negative since β(λ) > 0forλ<λF and β(λ) < 0forλ>λF . For large values of µ we have that λ → λF independently on the initial value λs. For this reason λF is referred to as a ultraviolet (UV) ﬁxed point (see Fig. 6.4.2)). The large scale (or small distance) behaviour of a ﬁeld theory of . λ

~~λ F~~

~~µ µ S~~

~~Fig. 6.4.2 - The behaviour of λ close to a UV ﬁxed point.~~

this type would be dictated by the value of λF . In particular if λF 1and λs <λF we would be always in the perturbative regime. Otherwise, starting from λs >λF the theory would become perturbative at large scales (see Fig. 6.4.1). About this point, notice that λ =0isaﬁxedpointsinceβ(0) = 0. However this is an infrared (IR) ﬁxed point, since the coupling goes to zero for

163 µ → 0 independently on the initial value λs. We say also that the fixed point is repulsive since λ goes away from the fixed point for increasing µ,whereasa UV fixed point is said to be attractive.

• Let us consider now the case of β(λ) < 0 for small values of λ and decreasing monotonically. For instance, suppose β(λ)=−aλ2 with a>0. Integrating the equation for the running coupling we get 1 λ(µ)=λs µ (6.88) s 1+aλ log µs In this case λ is a decreasing function of µ and goes to zero for µ →∞(that is λ = 0 is a UV ﬁxed point). This property is known as ”asymptoyic freedom” and it holds for non-abelian gauge theories that we will describe later on in the course. these are the only 4-dimensional theories to share this property. In higher dimensions other theories enjoy this property. For instance in d =6 the theory λϕ3 is asymptotically free. Notice that also this time the running coupling has a pole, but at a smaller scale than µs 1 µ − = e aλs (6.89) µs Therefore the coupling increases at large distances.

• Finally let us consider the case of β(λ) < 0 for small values of λ, becomes zero at λF andthenpositive.Thenβ (λF ) > 0. By expanding β(λ) around λF we get (by the same analysis we did before)

~~β (λF ) λ − λF µ = (6.90) λs − λF µs~~

~~and we see that λ → λF µ → 0 in a way independent on λs, whereas it goes away from λs for increasing µ. Therefore λF is an IR ﬁxed point (see Fig. 6.4.3)).~~

Let us now derive the perturbative expressions for γm and γd using the same method followed for β(λ). `e possibile ricavare. We diﬀerentiate the eq. (6.13) with respect to µ (notice that in this renormalization scheme b0 =1) ∞ ∞ ∂m bk ∂λ b 0=2m 1+ + m2 k (6.91) ∂µ k ∂µ k k=1 k=1 from which ∞ ∞ bk bk 0=2γm 1+ + β(λ) (6.92) k k k=1 k=1

~~164 . λ~~

~~λ F~~

~~µ µ S Fig. 6.4.3 - The behaviour of λ close to an IR fixed point. wherewehavemadeuseofthedefinitionγm =(µ/m)(∂m/∂µ). Using eq. (6.46) we find~~

~~db1 2γm − λ =0 dλ dbk d dbk+1 2γmbk − 1 − λ a − λ (6.93) dλ dλ 1 dλ or 1 db1 γm = λ (6.94) 2 dλ and db1 dbk d dbk+1 λ bk − 1 − λ a − λ = 0 (6.95) dλ dλ dλ 1 dλ~~

~~Using the eq. (6.24) for b1 we get for γm 2 λ 5 λ 3 γm(λ)= − + O(λ ) (6.96) 32π2 12 16π2~~

~~Let us now consider Zϕ (recall that although this quantity is infinite in the limit → 0, γd is finite). By differentiating eq. (6.14) we find (c0 =1)~~

∞ ∞ 1 ∂Zϕ 1 dλ ck 1 ck Zϕγd = µ = µ = β(λ) (6.97) 2 ∂µ 2 dµ k 2 k k=1 k=1 or ∞ ∞ ck 1 ck 1+ γd = (A + B) (6.98) k 2 k k=1 k=1 Using again the expression for β(λ) (see eq. (6.46)

~~1 dc1 γd = − λ (6.99) 2 dλ~~

~~165 dck+1 dc1 dck d λ = λck − 1 − λ a (6.100) dλ dλ dλ dλ 1 At the lowest order, using eq. (6.27) we get 2 1 λ 3 γd = + O(λ ) (6.101) 12 16π2~~

To conclude this Section, let us study the behaviour of Γ(n) when the theory has a UV ﬁxed point at λ = λF . Since for large scales λ → λF we may suppose that γm and γd go to γm(λF )eγd(λF ) respectively. Then, integrating the following equation (see eq. (6.61)) dm s = m(γm(λF ) − 1) (6.102) ds we have m(s)=m(1)sγm(λF )−1 (6.103) Analogously s ds γd(λF )=γd(λF )logs (6.104) 1 s Then, for large values of s (see eq. (6.59)

~~(n) (4−n−nγd(λF )) (n) γm(λF )−1 Γ (spi; m, λ, µ) → s Γ (pi; ms ,λF ,µ) (6.105)~~

Therefore, if 1 − γm(λF ) > 0 we can safely neglect the mass on the right hand side of this equation. From this equation we understand also the reason why γd is called the anomalous dimension. In non abelian gauge theories there is a UV ﬁxed point at λ =0whereγm =0 and therefore one can neglect the masses at large momenta. To justify why we have assumed that in the evaluation of γm and γd the only important contribution comes from λ = λF we notice that the integral

~~s ds γm(λ(s )) (6.106) 1 s can be rewritten (using ds/s = dλ/β) λ(s) γm(λ ) dλ (6.107) λ β(λ )~~

~~Therefore, barring the case in which γm and β have simultaneous zeroes, the integral is dominated by the points where there is a zero of the β function.~~

~~166 6.5 The coeﬃcients of the renormalization group equations and the renormalization conditions~~

We have already noticed that the coefficients of the renormalization group equations depend on the subtraction procedure. In particular, using the first two renormalization conditions considered in Section 6.2 a mass dependence of the coefficients is introduced. To study better this problem, let us consider, for generic renormalization conditions, the expression (6.12) for the bare coupling ∞ ak λB = µ a + (6.108) 0 k k=1 Differentiating with respect to µ we find ∞ ∞ ∞ ak ∂λ a ∂(m/µ) a˙ k a + + µ a + k + µ a˙ + = 0 (6.109) 0 k ∂µ 0 k ∂µ 0 k k=1 k=1 k=1 where the dot stands for the derivative with respect to m/µ. Notice also that (see eq. (6.5) ∂ m m µ = (γm − 1) (6.110) ∂µ µ µ With a familiar procedure, we put β = A + B, obtaining

~~a0 + Ba0 = 0 (6.111)~~

m a + Aa + Ba + (γm − 1)˙a = 0 (6.112) 1 0 1 µ 0 At the second order in λ − 3λ 3λ 1 3λ B = −λ 1+ G 1+ G ≈−λ 1 − G (6.113) 32π2 1 16π2 1 32π2 1 and 2 2 3λ 3λ 3λ m 3λ Aa = − + λ 1 − G − (γm − 1) G˙ (6.114) 0 16π2 32π2 1 8π2 µ 32π2 1 Since the right hand side is at least of the second order in λ, neglecting higher order terms we get 3λ2 m 3λ2 A = + G˙ (6.115) 16π2 µ 32π2 1 and for → 0 3λ2 m 3λ2 β = + G˙ (6.116) 16π2 µ 32π2 1 We see directly how the β function depends on the subtraction procedure. Similar expression can be obtained for γm e γd. Therefore, when using subtraction procedure

167 such that the coeﬃcients of the renormalization group equations depend explicitly on the masses, it is convenient to look for solutions in a region of momenta where the masses can be neglected. Another possibility is to choose the arbitrary scale µ such that m/µ can be neglected. Of course, changing the renormalization procedure will induce a ﬁnite renormalization in the couplings. For instance we get expressions of the type m λ = f λ, = λ + O(λ2) (6.117) µ

~~Therefore we obtain ∂λ ∂λ ∂(m/µ) m β = µ = µ f + µ f˙ = βf + (γm − 1)f˙ (6.118) ∂µ ∂µ ∂µ µ Working in a region where we can neglect the mass we have~~

~~β(λ)=β(λ)f (λ) (6.119)~~

and we see that a fixed point at λ = λF implies a fixed point in λ . Therefore the presence of a fixed point does not depend on the scheme used (at least in this approximation). It is possible to show that also the sign of the β function is scheme- independent.

~~168 Chapter 7~~

~~The path integral for Fermi ﬁelds~~

~~7.1 Fermionic oscillators and Grassmann algebras~~

The Feynman’s formulation of the path integral can be extended to ﬁeld theories for fermions. However, since the path integral is strictly related to the classical description of the theory, it could be nice to have such a ”classical” description for fermionic systems. In fact it turns out that it is possible to have a ”classical” (called pseudoclassical) description in a way such that its quantization leads naturally to anticommutation relations. In this theory one introduces anticommuting variables already at the classical level. This type of mathematical objects form the so called Grassmann algebras. Before proceeding further let us introduce some concepts about these algebras which will turn out useful in the following. First of all we recall that an algebra V is nothing but a vector space where it is deﬁned a bilinear mapping V × V → V (the algebra product). A Grassmann n algebra Gn is a vector space of dimensions 2 and it can be described in terms of n generators θi (i =1, ···,n) satisfying the relations

≡ ··· θiθj + θjθi [θi,θj]+ =0,i,j=1, ,n (7.1) Then, the elements constructed by all the possible products among the generators form a basis of Gn: 1,θi,θiθj,θiθjθk, ···,θ1θ2 ···θn (7.2) Notice that there are no other possible elements since

2 θi =0 (7.3) from eq. (7.1). The subset of elements of Gn generated by the product of an even (0) number of generators is called the even part of the algebra, Gn , whereas the subset (1) generated by the product of an odd number of generators is the odd part, Gn .It follows (0) (1) Gn = Gn ⊕Gn (7.4)

~~169 A decomposition of this kind is called a grading of the vector space, if it is possible to assign a grade to each element of the subspaces. In this case the grading is as follows~~

~~(0) 1. If x ∈Gn ,deg(x)=+1.~~

~~(1) 2. If x ∈Gn ,deg(x)=−1. Notice that this grading is also consistent with the algebra product, since if x, y are homogeneous elements of Gn (that is to say even or odd), then~~

~~deg(xy)=deg(x)deg(y) (7.5)~~

Since we have no time to enter into the details of the classical mechanics of the Grassmann variables, we will formulate the path integral starting directly from quantum mechanics. Let us recall that in the Feynman’s formulation one introduces states which are eigenstates of a complete set of observables describing the configuration space of the system. In the case of field theories one takes eigenstates of the field operators in the Heisenberg representation

~~φ(x, t)|ϕ(x),t = ϕ(x)|ϕ(x),t (7.6) this is possible since the ﬁelds commute at equal times~~

[φ(x, t),φ(y, t)]− =0 (7.7) and as a consequence it is possible to diagonalize simultaneously at fixed time the set of operators φ(x, t). For Fermi fields such a possibility does not arise since they anticommute at equal times. Therefore it is not possible to define simultaneous eigenstates of Fermi fields. To gain a better understanding of this point let us stay for the moment with free fields, and let us recall that this is nothing but a set of harmonic oscillators. This is true also for Fermi fields, except for the fact that these oscillators are of Fermi type, that is they are described by creation and annihilation operators satisfying the anticommutation relations † † † [ai,aj]+ = ai ,aj =0, ai,aj = δij (7.8) + + For greater simplicity take the case of a single Fermi oscillator, and let us try to diagonalize the corresponding annihilation operator

~~a|θ = θ|θ (7.9)~~

Then, from a2 = 0, as it foillows from eqs. (7.8), we must have θ2 = 0. Therefore the eigenvalue θ cannot be an ordinary number. However it can be thought as an odd element of a Grassmann algebra. If we have n Fermi oscillators and we require

~~ai|θ1, ···,θn = θi|θ1, ···,θn (7.10)~~

~~170 we get ai (aj|θ1, ···,θn )=θj (ai|θ1, ···,θn )=θjθi|θ1, ···,θn (7.11)~~

~~But aiaj = −ajai and therefore~~

~~[θi,θj]+ = 0 (7.12)~~

We see that the equations (7.10) are meaningful if we identify the eigenvalues θi with the generators of a Grassmann algebra Gn. In this case the equation (7.9) can be solved easily † |θ = eθa |0 (7.13) where |0 is the ground state of the oscillator. In fact

a|θ = a(1 + θa†)|0 = θaa†|0 = θ(1 − a†a)|0 = θ|0 = θ(1 + θa†|0 = θ|θ (7.14) wherewehaveused † eθa =1+θa† (7.15) since θ2 = a2 =0.Wehavealso ∂ ∂ a†|θ = a†(1 + θa†)|0 = a†|0 = (1 + θa†)|0 = |θ (7.16) ∂θ ∂θ

~~Here we have used the left-derivative on the Gn deﬁned by~~

~~∂ ··· ··· − ··· θi1 θi2 θik = δi1,iθi2 θi3 θik δi2,iθi1 θi3 θik ∂θi ··· − k−1 ··· + +( 1) δik,iθi1 θi3 θik−1 (7.17)~~

Suppose we want to define a scalar product in the vector space of the eigenstates of a (notice that these are vector spaces with coefficients in a Grassmann algebra. mathematically such an object is called a modulus over the algebra). Then it is convenient to extend the algebra G1 generated by θ to a comples Grassmann algebra, ∗ G2, generated by θ and θ . The operation ” ” is defined as the complex conjugation on the ordinary numbers and it is an involution (automorphism of the algebra with square one) with the property ··· ∗ ∗ ··· ∗ ∗ (θi1 θi2 θik ) = θik θi2 θi1 (7.18)

~~In the algebra G2 we can deﬁne normalized eigenstates~~

~~† |θ = eθa |0 N(θ, θ∗) (7.19)~~

~~∗ ∈G(0) where N(θ, θ ) 2 and the phase is ﬁxed by the convention N ∗(θ, θ∗)=N(θ, θ∗) (7.20)~~

~~171 The corresponding bra will be ∗ θ| = N(θ, θ∗) 0|eθ a (7.21) the normalization factor N(θ, θ∗) is determined by the condition~~

~~It follows ∗ N(θ, θ∗)=eθθ /2 (7.23) Let us recall some simple properties of a Fermi oscillator. the occupation number operator has eigenvalues 1 and 0. In fact~~

(a†a)2 = a†aa†a = a†(1 − a†a)a = a†a (7.24) from which a†a(a†a − 1) = 0 (7.25) Starting from the state with zero eigenvalue for a†a, we can build up only the eigenstate with eigenvalue 1, a†|0 ,sincea†2|0 = 0. Therefore the space of the eigenstates of a†a is a two-dimensional space. By deﬁning the ground state as 0 |0 = (7.26) 1 one sees easily that 00 † 01 a = ,a= (7.27) 10 00 (one writes down a as the most general 2 × 2 matrix and then requires a|0 =0, 2 † a =0and a, a + = 1). The state with occupation number equal to 1 is 1 |1 = (7.28) 0

~~In this basis the state |θ is given by~~

~~∗ † ∗ | θθ /2 θa | θθ /2 † | θ = e e 0 = e (1 + θa )0 ∗ ∗ θ = eθθ /2 (|0 + θ|1 )=eθθ /2 (7.29) 1 whereas ∗ θ| = eθθ /2 [ θ∗ 1 ] (7.30)~~

172 We see that in order to introduce eigenstates of the operator a has forced us to generalize the ordinary notion of vector space on the real or on the complex numbers, to a vector space on a Grassmann algebra (a modulus). In this space the linear combinations of the vectors are taken with coeﬃcients in G2 (or more generally in G2n). Let us now come to the deﬁnition of the integral over a Grassmann algebra. To this end we recall that the crucial element in order to derive the path integral formalism from quantum mechanics is the completeness relation for the eigenstates of the position operator dq|q q| = 1 (7.31)

What we are trying to do here is to think to the operators a as generalized position operators, and therefore we would like its eigenstates to satisfy a completeness relation of the previous type. Therefore we will ”deﬁne” the integration over the Grassmann algebra G2 by requiring that the states deﬁned in eqs. (7.29) and (7.30) satisfy ∗ 10 dθ dθ|θ θ| = ≡ 1 (7.32) 01 2 where the identity on the right hand side must be the identity on the two-dimensional vector space. Using the explicit representation for the states ∗ ∗ ∗ θ dθ dθ|θ θ| = dθ dθeθθ ( θ∗ 1) 1 ∗ ∗ ∗ θθ θ = dθ dθeθθ θ∗ 1 ∗ ∗ θθ θ 10 = dθ dθ = (7.33) θ∗ 1+θθ∗ 01 Therefore the integration must satisfy dθ∗dθ · θθ∗ = 1 (7.34) dθ∗dθ · θ = dθ∗dθ · θ∗ = dθ∗dθ · 1 = 0 (7.35)

These relations determine the integration in a unique way, since they ﬁx the value of the integral over all the basis elements of G2. The integral over a generic element of the algebra is then deﬁned by linearity. the previous integration rules can be rewritten as integration rules over the algebra G1 requiring ∗ ∗ ∗ ∗ ∗ [dθ, dθ ]+ =[θ, dθ ]+ =[θ, dθ]+ =[θ ,dθ]+ =[θ ,dθ ]+ = 0 (7.36) Then, we can reproduce the rules (7.34) and (7.35) by requiring dθ · θ =1, dθ · 1 = 0 (7.37)

~~173 For n Fermi oscillators we get n n ∗ † θiθi /2 θiai i=1 i=1 |θ1, ···,θn = e e |0 (7.38) with n ∗ dθi dθi |θ1, ···,θn θ1, ···,θn| = 1 (7.39) i=1 and integration rules~~

~~dθiθj = δij (7.40)~~

~~Assuming also~~

~~[dθi,θj]+ =[dθi,dθj]+ = 0 (7.41) Notice that the measure dθ is translationally invariant. In fact if f(θ)=a + bθ and ∈G(1) α 1 dθf(θ)= dθf(θ + α) (7.42) since’ dθbα = 0 (7.43)~~

~~7.2 Integration over Grassmann variables~~

Let us consider the Grassmann algebra Gn. Its generic element can be expanded over the basis elements as ··· ··· ··· ··· f(θ1, ,θn)=f0 + f1(i1)θi1 + f2(11.i2)θi1 θi2 + fn(i1, ,in)θi1 θin i1 i1,12 i1,···,in (7.44) Of course the coeﬃcients of the expansion are antisymmetric. If we integrate f over the algebra Gn only the last element of the expansion gives a non vanishing contribution. Noticing also that for the antisymmetry we have ··· ··· ··· ··· fn(i1, ,in)θi1 θin = n!fn(1, ,n)θ1 θn (7.45) i1,···,in it follows

dθn ···dθ1f(θ1, ···,θn)=n!fn(1, ···,n) (7.46) An interesting property of the integration rules for Grassmann variables is the integration by parts formula. This can be seen either by the observation that a derivative destroys a factor θi in each element of the basis and therefore ∂ dθn ···dθ1 f(θ1, ···,θn) = 0 (7.47) ∂θj

174 either by the property that the integral is invariant under translations (see the previous Section and Section 4.7). Let us now consider the transformation properties of the measure under a change of variable. For simplicity consider a linear transformation θi = aikθk (7.48) k The eq. (7.40) must be required to hold in each basis, since it can be seen easily that the previous transformation leaves invariant the multiplication rules. In fact,

~~θiθj = cikcjhθkθh = −cikcjhθhθk == −cjhcikθhθk = −θj θi (7.49) where cij is the inverse of the matrix aij. Therefore we require dθiθj = dθiθj = δij (7.50)~~

By putting dθi = dθkbki (7.51) k we obtain dθiθj = dθkbkiajhθh = bkiajhδhk =(ab)ij = δij (7.52) h,k h,k from which −1 bij =(a )ij (7.53) The transformation properties of the measure are ··· | −1| ··· dθn dθ1 =deta dθn dθ1 (7.54) where we have used the anticommutation properties of dθi among themselves. We see also that ∂θ det|a| =det (7.55) ∂θ and −1 ∂θ dθn ···dθ =det dθn ···dθ1 (7.56) 1 ∂θ Therefore the rule for a linear change of variables is −1 ∂θ dθn ···dθ f(θi)= dθn ···dθ1det f(θi(θi)) (7.57) 1 ∂θ This rule should be compared with the one for commuting variables ∂x dx ···dxnf(xi)= dx1 ···dxndet f(xi(xi)) (7.58) 1 ∂x

175 We see that the Grassmann measure transforms according to the inverse jacobian rather than with the jacobian itself. As in the commuting case, the more important integral to be evaluated for the path integral approach is the gaussian one Aijθiθj/2 T i,j θ Aθ/2 I = dθn ···dθ1e ≡ dθn ···dθ1e (7.59)

The matrix A is supposed to be real antisymmetric AT = −A (7.60) As shown in the Appendix a real and antisymmetric matrix can be reduced to the form ⎡ ⎤ 0 λ1 00··· ⎢ ⎥ ⎢ −λ1 000···⎥ ⎢ ⎥ ⎢ 000λ2 ···⎥ ⎢ ⎥ As = ⎢ 00−λ2 0 ···⎥ (7.61) ⎢ ⎥ ⎢ ·······⎥ ⎣ ·······⎦ ······· by means of an orthogonal transformation S. Then, let us do the following change of variables T θi = (S )ijθj (7.62) j we get T T T T T T θ Aθ =(S θ )AS θ = θ SAS θ = θ Asθ (7.63) and using det|S| =1 T ··· θ Asθ /2 I = dθn dθ1e (7.64) The expression in the exponent is given by

1 T θ Asθ = λ θ θ + ···+ λ n θn− θn, for n even 2 1 1 2 2 1 1 T θ Asθ = λ1θ θ + ···+ λ n−1 θn− θn− , for n odd (7.65) 2 1 2 2 2 1 From eq. (7.46) we get contribution to the integral only from the term in the T n expansion of the exponential (θ Asθ ) 2 for n even. In the case of n odd we get T I =0sinceθ Asθ does not contain θn. Therefore, for n even T n θ Asθ /2 1 1 T I = dθ ···dθ e = dθ ···dθ ( θ Asθ ) 2 n 1 n 1 ( n )! 2 2 1 n = dθ ···dθ ( )!λ ···λ n θ ···θ n 1 ( n )! 2 1 2 1 n √ 2 ··· n = λ1 λ 2 = detA (7.66)

~~176 wherewehaveuseddet|As| =det|A|.Then √ θT Aθ/2 dθn ···dθ1e = detA (7.67)~~

This result is valid also for n odd since in this case det|A| = 0. A similar result holds for commuting variables. In fact, let us deﬁne −xT Ax/2 J = dx1 ···dxne (7.68) with A a real and symmetric matrix. A can be diagonalized by an orthogonal transformation with the result T n −x Adx /2 1 J = dx ···dxne =(2π) 2 √ (7.69) 1 detA Again, the result for fermionic variables is opposite to the result of the commuting case. Another useful integral is θT Aθ/2+χT θ I(A; χ)= dθn ···dθ1e (7.70) with χi Grassmann variables. To evaluate this integral let us put

~~θ = θ + A−1χ (7.71)~~

~~We get~~

1 1 T θT Aθ + χT θ = (θ − χT A−1)A(θ + A−1χ)+χT (θ + A−1χ) 2 2 1 T 1 = θ Aθ + χT A−1χ (7.72) 2 2 T where we have used the antisymmetry of A and χT θ = −θ χ. Since the measure is invariant under translations we ﬁnd T T −1 T −1 √ ··· θ Aθ /2+χ A χ/2 χ A χ/2 I(A; χ)= dθn dθ1e = e detA (7.73) again to be compared with the result for commuting variables

n −xT Ax/2+J T x J T A−1J/2 (2π) 2 I(A; J)= dx1 ···dxne = e √ (7.74) detA The previous equations can be generalized to complex variables † ∗ ··· ∗ θ Aθ | | dθ1dθ1 dθndθne =detA (7.75)

177 and † † † † −1 ∗ ··· ∗ θ Aθ + χ θ + θ χ −χ A χ | | dθ1dθ1 dθndθne = e det A (7.76) The second of these equations follows from the ﬁrst one as in the real case. we will prove the ﬁrst one for the case n = 1. Starting from θT Aθ/2 dθ2dθ1e = A12 (7.77) and putting 1 ∗ i ∗ θ1 = √ (η + η ),θ2 = −√ (η − η ) (7.78) 2 2 we get 1 θT Aθ = −iη∗A η (7.79) 2 12 and recalling that (remember that here one has to use the inverse of the jacobian)

~~∗ dθ2dθ1 = idηdη (7.80) we get T ∗ θ Aθ/2 ∗ −iη A12η dθ2dθ1e = i dηdη e (7.81) and ∗ ∗ η (−iA12)η dηdη e = −iA12 (7.82)~~

~~7.3 The path integral for the fermionic theories~~

In this Section we will discuss the path integral formulation for a system described by anticommuting variables. Considering a Fermi oscillator, one could start from the ordinary quantum mechanical description and then by using the completeness in terms of Grassmann variable, it would be possible to derive the corresponding path integral formulation. We will give here only the result of this analysis. It turns out that the amplitude among eigenstates of the Fermi annihilation operator is given by

~~∗ ∗ θ (t )=θ ,t ∗ ∗ θ,t|θ, t = D(θ, θ∗)eθ θ(t )/2+θ (t)θ/2+iS (7.83) θ(t)=θ,t where D(θ, θ∗)= dθdθ∗ (7.84)~~

~~and t i S = (θ∗θ˙ − θ˙∗θ) − H(θ, θ∗) dt (7.85) t 2~~

178 where, for the Fermi oscillator H = θ∗θ. The expression is very similar to the one obtained in the commuting case if we realize that the previous formula is the analogue in the phase space using complex coordinates of the type q ± ip. It should be noticed the particular boundary conditions for the θ variables originating from the fact that the equations of motion are of the first order. Correspondingly we can assign only the coordinates at a particular time. In any case, since at the end we are interested in field theory where only the generating functional is relevant and where the boundary conditions are arbitrary (for instance we can take the fields vanishing at t = ±∞), and in any case do not affect the dependence of the generating functional on the external source. The extension to the Fermi field is done exactly as we have done in the bosonic case, and correspondingly the generating functional for the free Dirac theory will be i [ψ¯(i∂/ − m)ψ +¯ηψ + ψη¯ ] Z[η, η¯]= D(ψ, ψ¯)e (7.86) where η andη ¯ are the external sources (of Grassmann character). Using the integration formula (7.73) we get formally − − i − 1 (iη¯) / (iη) iη¯ / η Z[η, η¯]=Z[0]e i∂ − m = Z[0]e i∂ − m (7.87) where the determinant of the Dirac operator has been included in the term at zero source. In this expression we have to specify how to take the inverse operator. We know that this is nothing but the Dirac propagator defined by 4 (i∂/ − m)SF (x − y)=δ (x − y) (7.88) It follows 4 4 −i d xd yη¯(x)SF (x − y)η(y) Z[η, η¯]=Z[0]e (7.89) with 4 4 d k 1 −ikx d k /k + m −ikx SF (x) = lim e = lim e →0+ (2π)4 /k − m + i →0+ (2π)4 k2 − m2 + i (7.90) The Green’s functions are obtained by differentiating the generating functional with respect to the external sources (Grassmann differentiation). In particular, the two point function is given by 1 δZ 1 = 0|T (ψ(x)ψ¯(y))|0 = iSF (x − y) (7.91) Z[0] δη¯(x)δη(y) η=¯η=0 0|0 The T -product in this expression is the one for the Fermi fields, as it follows from the derivation made in the bosonic case, taking into account that the classical field of this case are anticommuting. We will not insist here further on this discussion, but all we have done in the bosonic case can be easily generalized for Fermi fields.

~~179 Chapter 8~~

~~The quantization of the gauge ﬁelds~~

~~8.1 QED as a gauge theory~~

Many ﬁeld theories possess global symmetries. These are transformations leaving invariant the action of the system and are characterized by a certain number of parameters which are independent on the space-time point. As a prototype we can consider the free Dirac lagrangian ¯ / L0 = ψ(x)[i∂ − m]ψ(x) (8.1) which is invariant under the global phase transformation

~~− ψ(x) → ψ (x)=e iQαψ(x) (8.2)~~

If one has more than one field, Q is a diagonal matrix having as eigenvalues the ¯ charges of the different fields measured in unit e. For instance, a term as ψ2ψ1φ, with φ a scalar field, is invariant by choosing Q(ψ1)=Q(φ) = 1, and Q(ψ2)=2. This is a so called abelian symmetry since − − − − − e iαQe iβQ = e i(α + β)Q = e iβQe iαQ (8.3)

It is also referred to as a U(1) symmetry. The physical meaning of this invariance lies in the possibility of assigning the phase to the ﬁelds in an arbitrary way, without changing the observable quantities. This way of thinking is in some sort of contradic- tion with causality, since it requires to assign the phase of the ﬁelds simultaneously at all space-time points. It looks more physical to require the possibility of assigning the phase in an arbitrary way at each space-time point. This invariance, formulated by Weyl in 1929, was called gauge invariance. The free lagrangian (8.1) cannot be gauge invariant due to the derivative coming from the kinetic term. The idea is

~~180 simply to generalize the derivative ∂µ to a so called covariant derivative Dµ having the property that Dµψ transforms as ψ,thatis~~

~~−iQα(x) Dµψ(x) → [Dµψ(x)] = e Dµψ(x) (8.4)~~

In this case the term ψD¯ µψ (8.5) will be invariant as the mass term under the local phase transformation. To construct the covariant derivative, we need to enlarge the field content of the theory, by introducing a vector field, the gauge field Aµ, in the following way

~~Dµ = ∂µ + ieQAµ (8.6)~~

~~The transformation law of Aµ is obtained from eq. (8.4)~~

[(∂µ + ieQAµ)ψ] =(∂µ + ieQAµ)ψ (x) −iQα(x) =(∂µ + ieQA )e ψ µ −iQα(x) 1 = e ∂µ + ieQ(A − ∂µα) ψ (8.7) µ e from which 1 A = Aµ + ∂µα (8.8) µ e The lagrangian density ¯ / ¯ µ ¯ µ Lψ = ψ[iD − m]ψ = ψ[iγ (∂µ + ieQAµ) − m)ψ = L0 − eψQγ ψAµ (8.9) is then invariant under gauge transformations, or under the local group U(1). We see also that by requiring local invariance we reproduce the electromagnetic interaction as obtained through the minimal substitution we discussed before. In order to determine the kinetic term for the vector ﬁeld Aµ we notice that eq. (8.4) implies that under a gauge transformation, the covariant derivative undergoes a unitary transformation

~~−iQα(x) iQα(x) Dµ → Dµ = e Dµe (8.10)~~

~~Then, also the commutator of two covariant derivatives~~

~~[Dµ,Dν]=[∂µ + ieQAµ,∂ν + ieQAν ]=ieQFµν (8.11) with Fµν = ∂µAν − ∂νAµ (8.12) transforms in the same way −iQα(x) iQα(x) Fµν → e Fµν e = Fµν (8.13)~~

~~181 The last equality follows from the commutativity of Fµν with the phase factor. The complete lagrangian density is then~~

µ 1 µν L = Lψ + LA = ψ¯[iγ (∂µ + ieQAµ) − m]ψ − Fµν F (8.14) 4 The gauge principle has automatically generated an interaction between the gauge field and the charged field. We notice also that gauge invariance prevents any 1 2 µ mass term, 2 M A Aµ. Therefore, the photon field is massless. Also, since the local invariance implies the global ones, by using the Noether’s theorem we find the conserved current as ∂L jµ = δψ = ψγ¯ µ(Qα)ψ (8.15) ∂ψ,µ from which, eliminating the infinitesimal parameter α,

~~Jµ = ψγ¯ µQψ (8.16)~~

~~8.2 Non-abelian gauge theories~~

The approach of the previous section can be easily extended to local non-abelian symmetries. We will consider the case of N Dirac ﬁelds. The free lagrangian N ¯ / L0 = ψa(i∂ − m)ψa (8.17) a=1 is invariant under the global transformation

Ψ(x) → Ψ(x)=AΨ(x) (8.18) where A is a unitary N × N matrix, and we have denoted by Ψ the column vector with components ψa. In a more general situation the actual symmetry could be a subgroup of U(N). For instance, when the masses are not all equal. So we will consider here the gauging of a subgroup G of U(N). The fields ψa(x) will belong, in general, to some reducible representation of G.DenotingbyU the generic element of G, we will write the corresponding matrix Uab acting upon the fields ψa as − A U = e iαAT ,U∈ G (8.19) where T A denote the generators of the Lie algebra associated to G,Lie(G), (that is the vector space spanned by the infinitesimal generators of the group) in the fermion representation. The generators T A satisfy the algebra

~~A B AB C [T ,T ]=ifC T (8.20)~~

182 AB where fC are the structure constants of Lie(G). For instance, if G = SU(2), and we take the fermions in the fundamental representation, ψ Ψ= 1 (8.21) ψ2 we have σA T A = ,A=1, 2, 3 (8.22) 2 where σA are the Pauli matrices. In the general case the T A’s are N × N hermitian matrices that we will choose normalized in such a way that 1 Tr(T AT B)= δAB (8.23) 2

To make local the transformation (8.19), means to promote the parameters αA to space-time functions αA → αA(x) (8.24) Notice, that now the group does not need to be abelian, and therefore, in general − A − A − A − A e iαAT e iβAT = e iβAT e iαAT (8.25)

~~Let us now proceed to the case of the local symmetry by deﬁning again the concept of covariant derivative~~

~~DµΨ(x) → [DµΨ(x)] = U(x)[Dµψ(x)] (8.26)~~

~~We will put again Dµ = ∂µ + igBµ (8.27) where Bµ is a N × N matrix acting upon Ψ(x). In components~~

~~µ µ µ Dab = δab∂ + ig(B )ab (8.28)~~

~~The eq. (8.26) implies~~

DµΨ → (∂µ + igBµ)U(x)Ψ −1 = U(x)∂µΨ+U(x)[U (x)igBµU(x)]Ψ + (∂µU(x))Ψ −1 −1 = U(x)[∂µ + U (x)igBµU(x)+U (x)∂µU(x)]Ψ (8.29) therefore −1 −1 U (x)igBµU(x)+U (x)∂µU(x)=igBµ (8.30) and −1 i −1 B (x)=U(x)Bµ(x)U (x)+ (∂µU(x))U (x) (8.31) µ g

~~183 For an inﬁnitesimal transformation~~

A U(x) ≈ 1 − iαA(x)T (8.32) we get A 1 A δBµ(x)=−iαA(x)[T ,Bµ(x)] + (∂µαA(x))T (8.33) g A Since Bµ(x) acquires a term proportional to T , the transformation law is consistent with a Bµ linear in the generators of the Lie algebra, that is

~~µ µ A (B )ab ≡ AA(T )ab (8.34)~~

~~The transformation law for Aµ becomes~~

µ AB µ 1 µ δA = f αAA + ∂ αC (8.35) C C B g The difference with respect to the abelian case is that the field undergoes also a homogeneous transformation. The kinetic term for the gauge fields is constructed as in the abelian case. In fact the quantity [Dµ,Dν]Ψ ≡ igFµν Ψ (8.36) in virtue of the eq. (8.26), transforms as Ψ under gauge transformations, that is

~~([Dµ,Dν]Ψ) = igFµν Ψ = igFµν U(x)Ψ = U(x)([Dµ,Dν]Ψ) = U(x)(igFµν ) Ψ (8.37)~~

~~This time the tensor Fµν is not invariant but transforms homogeneously, since it does not commute with the gauge transformation as in the abelian case~~

~~−1 Fµν = U(x)Fµν U (x) (8.38)~~

~~The invariant kinetic term will be assumed as~~

~~1 µν LA = − Tr[Fµν F ] (8.39) 2~~

~~Let us now evaluate Fµν~~

~~igFµν =[Dµ,Dν]=[∂µ + igBµ,∂ν + igBν] 2 = ig(∂µBν − ∂ν Bµ) − g [Bµ,Bν] (8.40) or Fµν =(∂µBν − ∂ν Bµ)+ig[Bµ,Bν] (8.41) in components µν µν C F = FC T (8.42)~~

184 with µν µ ν ν µ AB µ ν FC = ∂ AC − ∂ AC − gfC AAAB (8.43) The main feature of the non-abelian gauge theories is the bilinear term in the pre- AB vious expression. Such a term comes because fC = 0, expressing the fact that G is not abelian. The kinetic term for the gauge ﬁeld, expressed in components, is given by 1 µν LA = − FµνAFA (8.44) 4 A

Therefore, whereas in the abelian case LA is a free lagrangian (it contains only quadratic terms), now it contains interaction terms cubic and quartic in the fields. The physical motivation lies in the fact that the gauge fields couple to everything which transforms in a non trivial way under the gauge group. Therefore they couple also to themselves (remember the homogeneous piece of transformation). To derive the equations of motion for the gauge fields, let us consider the total action 4 µ d x Ψ(¯ i∂/ − m)Ψ − gΨ¯ γµB Ψ + SA (8.45) V where 1 4 µν SA = − d xTr(Fµν F ) (8.46) 2 V and the variation of SA 4 µν δSA = − d xT r(Fµν δF ) (8.47) V

~~Using the deﬁnition (8.41) for the ﬁeld strength we get~~

δFµν = ∂µδBν + ig(δBµ)Bν + igBµδ(Bν) − (µ ↔ ν) (8.48) from which 4 µν δSA = −2 d xT r[F (∂µδBν + ig(δBµ)Bν + igBµ(δBν))] (8.49) V where we have taken into account the antisymmetry properties of Fµν . Integrating by parts we obtain 4 µν µν µν δSA = −2 d xT r[−(∂µF )δBν − igBµF δBν + igF BµδBν] V 4 µν µν =2 d xT r [(∂µF + ig[Bµ,F ]) δBν] V 4 µν µν = d x (∂µF + ig[Bµ,F ])A δAνA (8.50) V

~~185 where we used the cyclic property of the trace. By taking into account also the free term for the Dirac ﬁelds and the interaction we ﬁnd the equations of motion~~

~~µνA µν A ν A ∂µF + ig[Bµ,F ] = gΨ¯ γ T Ψ µ (i∂/ − m)Ψ = gγµB Ψ (8.51)~~

~~A From the ﬁrst equation we see that the currents Ψ¯ γµT Ψ are not conserved. In fact the conserved currents turn out to be~~

~~A A µ A Jν = Ψ¯ γνT Ψ − i[B ,Fµν ] (8.52)~~

The reason is that under a global transformation of the symmetry group, the gauge fields are not invariant, said in different words they are charged fields with respect to the gauge fields. In fact we can verify immediately that the previous currents are precisely the Noether’s currents. Under a global variation we have

µ AB µ A δAC = fC αAAB,δΨ=−iαAT Ψ (8.53) and we get µ ∂L ∂L j = δΨ+ δAνC (8.54) ∂Ψ,µ ∂Aν,µC from which µ µ A µν AB j = Ψ¯ γ αAT Ψ − FC fC αAAνB (8.55) ABC AB In the case of simple compact Lie groups one can deﬁne f = fC with the property f ABC = f BCA. It follows

~~µν ABC A ν ν A A FC f AνBT = i[B ,Fνµ] ≡ i[B ,Fνµ] T (8.56) Therefore µ µ A νµ A j = Ψ¯ γ αAT Ψ − i[Bν,F ] αA (8.57)~~

After division by αA we get the Noether’s currents (8.52). The contribution of the gauge ﬁelds to the currents is also crucial in order they are conserved quantities. In fact, the divergence of the fermionic contribution is given by

µ A A µ µ A A µ ∂ (Ψ¯ γµT Ψ) = −igΨ¯ γµT B Ψ+igΨ¯ γµB T Ψ=−igΨ¯ γµ[T ,B ]Ψ (8.58) which vanishes for abelian gauge ﬁelds, whereas it is compensated by the gauge ﬁelds contribution in the non abelian case.

~~8.3 Path integral quantization of the gauge theories~~

~~As we have seen at the beginning of this course that the quantization of the electromagnetic ﬁeld is far from being trivial, the diﬃculties being related to the gauge~~

186 invariance of the theory. In this Section we would like to discuss how these difficul- ties arise also in the path integral quantization. To this end let us recall the general scheme of field quantization in this approach. First of all we will deal only with the perturbative approach, meaning that the only thing that one needs to evaluate is the generating functional for the free case. The free case is defined by the quadratic part of the action. Therefore the evaluation of the path integral reduces to calculate a Gaussian integral of the type given in eqs. (7.73) and (7.73) for the fermionic and bosonic case respectively. For instance, in the bosonic case one evaluates

n −xT Ax/2+J T x J T A−1J/2 (2π) 2 I(A; J)= dx1 ···dxne = e √ (8.59) detA In our case the operator A appearing in the gaussian factor is nothing but the wave operator. The result of the integration is meaningful only if the operator A is non- singular (it needs to have an inverse) otherwise the integral is not well deﬁned. This is precisely the point where we run into problems in the case of a gauge theory. Remember from eq. (1.132) that the wave operator for the electromagnetic ﬁeld in momentum space is given by

~~2 µν µ ν (−k g + k k )Aν (k) = 0 (8.60)~~

~~We see that the operator has a null eigenvector kν,since~~

~~2 µν µ ν (−k g + k k )kν = 0 (8.61) and therefore it is a singular operator. This is strictly related to the gauge invariance, which in momentum space reads~~

~~Aµ(k) → Aµ(k)+kµΛ(k) (8.62)~~

In fact, we see that the equations of motion are invariant under a gauge transformation precisely because the wave operator has kµ as a null eigenvector. Another way of seeing this problem is the following: in order to get physical results, as to evaluate S matrix elements, it is necessary to integrate over gauge-invariant functionals iS D(Aµ)F [Aµ]e (8.63)

Ω Ω with F [Aµ]=F [Aµ ](hereAµ is the gauge transformed of Aµ). But being the action and the functional F gauge invariant, it follows that the integrand is invariant along the gauge group orbits. The gauge group orbits are defined as the set of points, in the field space, which can be reached by a given Aµ via a gauge transformation. In Ω other words, given Aµ, its orbit is given by all the fields Aµ obtained by varying the parameters of the gauge group (that is varying Ω), see Fig. 8.3.1. For instance, in the abelian case Ω Aµ = Aµ + ∂µΩ (8.64)

~~187 .~~

~~Ω Aµ~~

~~Aµ Ω A'µ~~

~~A'µ~~

~~Fig. 8.3.1 - The orbits of the gauge group in the ﬁeld space.~~

Since the orbits define equivalence classes (or cosets) we can imagine the field space as the set of all the orbits. Correspondingly we can divide our functional integral in a part parallel and in one perpendicular to the orbits. Then, the reason why our functional integral is not well defined depends from the fact that the integrand along the space parallel to the orbits is infinite (the integrand being gauge invariant does not depend on the corresponding variables). However this observation suggests a simple solution to our problem. We could define the integral by dividing it by the integral along the part parallel to the orbit (notice that is the same as the volume of the gauge group). In order to gain a clear understanding of this problem let us consider the following very simple example in two dimensions 1 +∞ − a2(x − y)2 Z = dxdye 2 (8.65) −∞

This is as reducing the space-time two two points with x and y the possible values of the ﬁeld in those points. At the same time the exponent can be seen as an euclidean action with the (gauge) symmetry

~~xΩ = x +Ω,yΩ = y + Ω (8.66)~~

The integral (8.65) is infinite due to this invariance. In fact this implies that the integrand depends only on x − y, and therefore the integration over the remaining variable gives rise to an infinite result. In order to proceed along the lines we have described above, let us partition the space of the field, that is the two-dimensional space (x, y) in the gauge group orbits. Given a point (x0,y0), the group orbit is given by the set of points (xΩ = x +Ω,yΩ = y + Ω) corresponding to the line x − y = x0 − y0, as shown in Fig. 8.3.2 We can perform the integral by integrating

~~188 . y~~

~~ΩΩ (x , y )~~

~~(x , y ) 00 x~~

~~f(x,y) = 0~~

x + y = c Fig. 8.3.2 - The field space (x, y) partitioned in the orbits of the gauge group. We show also the line x + y = c and the arbitrary line f(x, y)=0(dashed). along the perpendicular space, the line x + y = c (c is a constant), and then along the parallel space. It is this last integral which gives a divergent result, since we are adding up the same contribution an infinite number of times. therefore the obvious way to do is to introduce as integration variables x − y and x + y, integrating only on x − y. Or, said in a different way, perform both integrals and dividing up by the integral over x + y, which cancels the divergence at the numerator. This can be done in a more systematic way by using the following identity +∞ 1 1= dΩδ (x + y)+Ω (8.67) −∞ 2

~~Then we multiply Z per by this factor obtaining ⎡ ⎤ 1 1 − a2(x − y)2 Z = dΩ ⎣ dxdyδ (x + y)+Ω e 2 ⎦ (8.68) 2~~

The integral inside the parenthesis is what we are looking for since it is evaluated along the line x+y = constant. Of course the result should not depend on the choice of this constant (it corresponds to translate the integration line in a way parallel to the orbits). In fact, by doing the change of variables

~~x → x +Ω,y→ y + Ω (8.69)~~

~~189 we get ⎡ ⎤ 1 1 − a2(x − y)2 Z = dΩ ⎣ dxdyδ (x + y) e 2 ⎦ (8.70) 2~~

As promised the inﬁnite factor is now factorized out and we see also clearly that this nothing but the volume of the gauge group (Ω parameterize the gauge group, which in the present case is isomorphic to R1). Then, we deﬁne our integral as 1 Z 1 − a2(x − y)2 Z = = dxdyδ (x + y) e 2 (8.71) VG 2 with VG = dΩ (8.72)

In this way the integral is deﬁned by integrating over the particular surface x+y =0. Of course we would get the same result integrating over any other line (or surface in the general case) having the property of intersecting each orbit only once. For instance, let us choose the line (see Fig. 8.3.2)

~~f(x, y) = 0 (8.73)~~

Then, proceeding as in eq. (8.67) we consider the following identity, which allows us to deﬁne a gauge invariant function ∆f (x, y) Ω Ω 1=∆f (x, y) dΩδ f(x ,y ) =∆f (x, y) dΩδ [f(x +Ω,y+ Ω)] (8.74)

In the previous case we had f(x, y)=(x + y)/2 = 0 giving rise to f(xΩ,yΩ)= (x + y)/2 + Ω = 0, from which ∆f (x, y) = 1. Multiplying Z by the expression (8.74) we get ⎡ ⎤ 1 − 2 − 2 ⎣ a (x y) ⎦ Z = dΩ dxdy∆f (x, y)δ [f(x +Ω,y+Ω)]e 2 (8.75)

Let us show that ∆f (x, y) is gauge invariant. In fact by the transformation x → x +Ω, y → y +Ω we obtain 1=∆f (x +Ω,y+Ω) dΩδ [f(x +Ω+Ω,y+Ω+Ω)] (8.76) and changing the integration variable Ω → Ω+Ω 1=∆f (x +Ω,y+Ω) dΩδ [f(x +Ω,y+ Ω)] (8.77)

~~190 Comparison with eq. (8.74) gives~~

~~∆f (x, y)=∆f (x +Ω,y+Ω) (8.78)~~

The integral inside the parenthesis in eq. (8.75) does not depend on Ω (that is is gauge invariant). This can be shown exactly as we did before in the case f(x, y)= (x + y)/2. Therefore we can again factorize the integration over Ω, being the sum of inﬁnite gauge copies ⎡ ⎤ 1 − 2 − 2 ⎣ a (x y) ⎦ Z = dΩ dxdy∆f (x, y)δ [f(x, y)] e 2 (8.79) and deﬁne −1 2 − 2 Z a (x y) Z = = dxdy∆f (x, y)δ [f(x, y)] e 2 (8.80) VG

The function ∆f (x, y) (called the Faddeev-Popov determinant) can be calculated directly from its deﬁnition. To this end notice that in (8.79), ∆f (x, y)appears multiplied by δ[f(x, y)]. As a consequence, we need to know ∆f only at points very close at the surface f(x, y) = 0. But in this case, the equation f(xΩ,yΩ)=0has the only solution Ω = 0, since by hypothesis the surface f(x, y) crosses the gauge orbits only once. Therefore we can write 1 δ[f(xΩ,yΩ)] = δ[Ω] (8.81) ∂fΩ

∂Ω Ω=0 giving rise to ∂fΩ ∆f (x, y)= (8.82) ∂Ω Ω=0 Ω Ω Ω where f = f(x ,y ). Clearly ∆f represents the answer of the gauge ﬁxing f =0 to an inﬁnitesimal gauge transformation. As a last remark, notice that our original integral can be written as 1 − xiAijxj Z = d2xe 2 (8.83) with 1 −1 A = a2 (8.84) −11 a singular matrix. As we have already noticed the singularity of A is related to the gauge invariance, and the solution we have given here solves also the problem of the singularity of A. In fact, integrating over x − y only, means to integrate only on the eigenvectors of A corresponding to non zero eigenvalues.

191 Since the solution we have found to our problem is of geometrical type, it is very easy to generalize it to the case of a gauge ﬁeld theory, abelian or not. We start by choosing a surface in the gauge ﬁeld space given by µ fB(AB)=0,B=1, ···,n (8.85) where n is the number of parameters of LieG. Also, this surface should intersect the gauge orbits only once. To write down the analogue of equation (8.74) we need also a measure on the gauge group. This can be obtained as follows. We have seen that we are interested in the solutions of

~~Ω fB(Aµ ) = 0 (8.86) around fB(Aµ) = 0, therefore it is enough to know the measure around the identity of the gauge group. In such a case the transformation Ω can be written as~~

~~A Ω ≈ 1+iαA(x)T (8.87) Since at the ﬁrst order in the group parameters~~

~~A ΩΩ ≈ 1+i(αA + αA)T (8.88) we see that the invariant measure of the group, around the identity, is given by dµ(Ω) = dαA(x) (8.89) A,x~~

~~It follows that for inﬁnitesimal transformations dµ(ΩΩ)=dµ(ΩΩ) = dµ(Ω) (8.90) next we deﬁne the Faddeev-Popov determinant as in (8.74) Ω 1=∆f [Aµ] dµ(Ω)δ[fA(Aµ )] (8.91)~~

In this expression we have introduced a functional delta-function, deﬁned by the identity 1= dµ(Ω)δ[Ω] (8.92) having the usual properties extended to the functional case. We can see that ∆f is a gauge invariant functional, since −1 Ω−1 (Ω−1Ω) ∆f [Aµ ]= dµ(Ω)δ[fA(Aµ ] (8.93) and changing variables ΩΩ=Ω we get −1 Ω−1 Ω Ω ∆f (aµ )= dµ(Ω Ω )δ[fA(Aµ )] = dµ(Ω)δ[fA(Aµ )] (8.94)

192 Consider now the naive path integral 4 i LAd x Z = D(Aµ)F [Aµ]e (8.95) where F is a gauge invariant functional. Since in the following F will not play any particular role we will discuss only the case F = 1, but all the following considerations are valid for an arbitrary gauge invariant F . We multiply Z by the identity (8.91) ⎡ ⎤ 4 i LAd x ⎢ Ω ⎥ Z = dµ(Ω) ⎣ D(Aµ)e δ[fA(Aµ )]∆f [Aµ]⎦ (8.96)

~~The functional measure is invariant under gauge transformations~~

−1 i −1 Aµ → ΩAµΩ + (∂µΩ)Ω (8.97) g since the homogeneous part has determinant one, and the inhomogeneous part gives Ω rise to a translation. Then by performing the change of variables Aµ → Aµ ,and using the gauge invariance of D(Aµ), LA and ∆f [Aµ], we get ⎡ ⎤ 4 ⎢ i LAd x ⎥ Z = dµ(Ω) ⎣ D(Aµ)e δ[fA(Aµ)]∆f [Aµ]⎦ (8.98)

~~Again, we deﬁne the functional integral as 4 Z Z i LAd x Z → = = D(Aµ)e δ[fA(Aµ)]∆f [Aµ] (8.99) VG dµ(Ω)~~

The evaluation of ∆f goes as before. Since in the previous equation it appears Ω multiplied by δ[fA(Aµ)], the only solution of fA(Aµ ) = 0 in this neighborhood is Ω = 1, or αA = 0. therefore Ω A Ω −1 δf (Aµ ) δ[fA(Aµ )] = det δ[αA] (8.100) δαB αA=0 from which Ω δfA(Aµ ) ∆f [Aµ]=det (8.101) δαB αA=0

Once again, the Faddeev-Popov determinant δf [Aµ] is the answer of the gauge-fixing to an infinitesimal gauge transformation. The final expression that we get for Z is Ω L 4 δfA(Aµ ) i Ad x Z = D(Aµ)δ[fA(Aµ)]det e (8.102) δαB αA=0

193 This expression gives the right measure to be used in the path integral quantization of the gauge fields. In particular, the vacuum expectation value (v.e.v.) of a gauge field functional will be defined as ∗ iS 0|T (F [A])|0 f = D(Aµ)δ[f]∆f e F [A] (8.103) where the index f specifies the gauge in which we evaluate the v.e.v. of F .About this point, we can prove an important result, that is: if F [A] is a gauge invariant functional F [AΩ]=F [A] (8.104) than its v.e.v. does not depend on the gauge fixing fA = 0. We will prove this result by showing the relation between the v.e.v.’s of F [A] in due different gauges. Consider the identity | ∗ | D iS Ω 0 T (F [A]) 0 f1 = (Aµ)δ[f1]∆f1 e F [A] dµ(Ω)δ[f2 ]∆f2 (8.105)

Ω ≡ Ω wherewehaveusedeq.(8.91)anddeﬁnedf2 [A] f2[A ]. By change of variable Ω Aµ → Aµ ,weget | ∗ | D Ω−1 iS Ω−1 0 T (F [A]) 0 f1 = dµ(Ω) (Aµ)δ[f1 ]∆f1 e F [A ]δ[f2]∆f2 D iS Ω Ω = dµ(Ω) (Aµ)e δ[f2]∆f2 δ[f1 ]∆f1 F [A ] (8.106) where in the second line we have made a further change of variables Ω → Ω−1 and used the invariance of dµ(Ω). Then | ∗ | | ∗ Ω Ω | 0 T (F [A]) 0 f1 = dµ(Ω) 0 T (δ[f1 ]∆f1 F [A ]) 0 f2 (8.107)

~~This is the relation we were looking for. If F is gauge invariant we get at once~~

~~∗ ∗ 0|T (F [A])|0 f1 = 0|T (F [A])|0 f2 (8.108)~~

For the following we will define a generating functional A µ i d4xη A A iS µ A Zf [ηµ ]= D(Aµ)δ[f]∆f e e (8.109) which is not gauge invariant, but in terms of which we can easily construct the v.e.v.’s of gauge invariant quantities, since it is built up in terms of the correct functional measure. For the purpose of building up the perturbative theory it is convenient to write the integrand of eq. (8.109) in the form exp(iSeff ), where Seff is an effective action

194 taking into account the gauge ﬁxing f = 0 and the Faddeev-Popov determinant. We will discuss later the exponentiation of ∆f . As far as δ[f] it is concerned, we can exponentiate it by using its Fourier representation, or, more conveniently by choosing the gauge condition in the form fA(A) − BA(x)=0,withBA(x)asetof arbitrary functions independent on the gauge ﬁelds. Then Ω Ω δ(fA[A ] − BA) δfA[A ] ∆(f−B)[A]=det =det =∆f [A] (8.110) δαB αA=0 δαB αA=0 and therefore Ω 1=∆f [A] dµ(Ω)δ[fA − BA] (8.111)

~~We can further multiply this identity by the following expression which does not depend on the gauge ﬁelds i − d4x B2 (x) 2β A constant = D(B)e A (8.112)~~

~~We get i 4 Ω 2 − d x (fA (x)) 2β A constant = ∆f [A] dµ(Ω)D(B)e δ(fA − BA) i 4 Ω 2 − d x (fA (x)) 2β A =∆f [A] dµ(Ω)e (8.113)~~

Multiplying this expression by the functional (8.95)we can show as before that the gauge volume factor factorizes, allowing us to deﬁne a class of generating functionals 4 1 2 i d x(LA − (fA) ) 4 A µ i d xηµ AA A 2β A Zβ[ηµ ]= D(Aµ)∆f [A]e e (8.114)

Finally, deﬁning 1 4 2 Sf = − d x (fA(x)) (8.115) 2β A we ﬁnd the analogue of eq. (8.107) is Ω ∗ ∗ iSf1 Ω 0|T (F [A])|0 f1 = dµ(Ω) 0|T (∆f1 e F [A ])|0 f2 (8.116) showing again that the v.e.v. a gauge invariant functional gives rise to a gauge invariant v.e.v. for the corresponding operator.

~~195 8.4 Path integral quantization of QED~~

~~Let us now discuss the path integral quantization of QED in a covariant gauge. We start from the lagrangian~~

~~µ 1 µν L = ψ¯ [iγ (∂µ + ieAµ) − m] ψ − Fµν F (8.117) 4 with Fµν = ∂µAν − ∂νAµ (8.118) invariant under~~

~~− ψ(x) → ψ (x)=e ieα(x)ψ(x) 1 Aµ(x) → A (x)=Aµ(x)+ ∂µα(x) (8.119) µ e The Lorentz gauge is deﬁned by~~

µ f(Aµ)=∂ Aµ(x) = 0 (8.120) and Ω µ 1 f(A )=∂ Aµ(x)+ α(x) (8.121) µ e with Ω δf(Aµ ) 1 4 = xδ (x − y) (8.122) δα(y) α=0 e Then the Faddeev-Popov determinant turns out to be ﬁeld-independent and it can be absorbed into the normalization of the generating functional, which is given by 4 µ 4 i d xL i [¯ηψ + ψη¯ + ηµA ]d x µ Z[η, η,¯ ηµ]=N D(ψ, ψ¯)D(Aµ)e δ[∂µA ]e (8.123) We extract the interaction term in the usual way 1 δ 1 δ 1 δ −ie d4x − γµ i δη i δηµ i δη¯ Z[η, η,¯ ηµ]=e Z0[η, η,¯ ηµ] (8.124) with 4 µ 4 i d xL i [¯ηψ + ψη¯ + ηµA ]d x ¯ 0 µ Z0[η, η,¯ ηµ]=N D(ψ, ψ)D(Aµ)e δ[∂µA ]e (8.125) Writing 1 µν 4 1 4 µν µ ν SA = − Fµν F d x = d xAµ(g − ∂ ∂ )Aν (8.126) 4 2

~~196 µ using δ[∂µA ] and integrating over the Fermi ﬁelds we get~~

~~−i η¯(x)SF (x − y)η(y) Z0[η, η,¯ ηµ]=e i µ Aµ(x)A (x) µ µ i ηµ(x)A (x) ·N D(Aµ)δ[∂µA ]e2 e (8.127)~~

~~µ Then we exponentiate the δ[∂µA ] through its Fourier transform µ µ µ i C(x)∂µA (x) −i ∂µC(x)A (x) δ[∂µA ]= D(C)e = D(C)e (8.128) obtaining the following expression for Z0~~

~~−i η¯(x)SF (x − y)η(y) Z0[η, η,¯ ηµ]=e 1 µ µ i Aµ(x)A (x)+(ηµ(x) − ∂µC(x))A (x) ·N D(Aµ)D(C)e 2 (8.129)~~

~~Doing the gaussian integral over Aµ we get~~

~~−i η¯(x)SF (x − y)η(y) Z0[η, η,¯ ηµ]=e 1 i µ µ i(ηµ(x) − ∂µC(x)) i(η (y) − ∂ C(y)) 2 ·N D(C)e x,y~~

~~i 1 µ − − − ηµ η = e i η¯(x)SF (x y)η(y) e 2 1 µ i 1 µ i ∂µC η − ∂µC ∂ C · D(C)e 2 (8.130)~~

~~After integrating by parts, we can rewrite the integral over C(x)as~~

~~1 µ i −i C ∂µη + CC D(C)e 2 (8.131) and doing again this gaussian integral we ﬁnd~~

~~i µ ∂µ∂ν ν 1 i i η η − µ − ν 2 2 ∂µη (+i) ∂ν η = e e2 (8.132)~~

~~The ﬁnal result is i gµν ∂µ∂ν − ηµ − ην −i η¯(x)SF (x − y)η(y) 2 2 Z0[η, η,¯ ηµ]=Z[0]e e (8.133)~~

~~197 Introducing the function 4 d k gµν kµkν −ikx Gµν = lim − + e (8.134) →0+ (2π)4 k2 + i (k2 + i)2 we write~~

i µ ν − η (x)Gµν (x − y)η (y) −i η¯(x)SF (x − y)η(y) Z0[η, η,¯ ηµ]=Z[0]e e 2 (8.135) The photon propagator in this gauge is given by 1 0|T (Aµ(x)Aν (x))|0 = iGµν (x − y) (8.136) 0|0 Notice that this propagator is transverse, that is it satisﬁes the Lorentz condition

~~µ ∂ Gµν (x) = 0 (8.137) The Feynman’s rules in this gauge are the same discussed in Section 1.5.3, except for th photon propagator which is gµν kµkν −i − (8.138) k2 + i (k2 + i)2~~

Of course, the additional term in the propagator proportional to kµkν does not affect the physics since the photon is coupled to a conserved current. We can perform an analogous calculation but using the gauge fixing in the form f − B discussed at the end of the previous Section, that is using the generating functional (8.114) 4 1 µ 2 i d x L− (∂µA ) 2β Zβ[η, η,¯ ηµ]=N D(ψ, ψ¯)D(Aµ)e 4 µ d x ηψ¯ + ψη¯ + ηµA · e (8.139) In this case the only effect of the gauge fixing is to change the lagrangian L to µ 2 L = L−(∂µA ) /(2β). Of course, the addition of this term removes the problem of the singular wave operator. In fact we can write 4 1 4 µ 1 ν d xL = d xA gµν − (1 − )∂µ∂ν A (8.140) 2 β Therefore we have to evaluate the functional integral 4 1 µ ν µ i d x A Kµν A + ηµA −i η¯(x)SF (x − y)η(y) 2 Z0[η, η,¯ ηµ]=Ne D(Aµ)e (8.141)

~~198 with β − 1 Kµν = gµν − ∂µ∂ν (8.142) β obtaining~~

~~i µν ν − ηµ(x)G (x − y)η (y) −i η¯(x)SF (x − y)η(y) Zβ[η, η,¯ ηµ]=Z[0]e e 2 (8.143) where νρ ρ 4 Kµν G (x)=δµδ (x) (8.144)~~

We can see that Kµν is non singular for β = ∞. Going to momentum space, eq. (8.144) gives 4 − 4 d k 2 β 1 νρ −ikx ρ d k −ikx −gµν k + kµkν G (k)e = δ e (8.145) (2π)4 β µ (2π)4 or − 2 β 1 νρ ρ −gµν k + kµkν G (k)=δ (8.146) β ν This can be solved by writing the most general second rank symmetric tensor function of kµ Gµν (k)=Agµν + Bkµkν (8.147) with the coeﬃcients such that β − 1 1 −Ak2 =1, A − Bk2 = 0 (8.148) β β Solving these equations

gµν kµkν Gµν (k)=− +(1− β) (8.149) k2 k4 The singularities are deﬁned by the usual i term, and we get 4 d k gµν kµkν −ikx Gµν (x) = lim − +(1− β) e (8.150) →0+ (2π)4 k2 + i (k2 + i)2

The discussion made at the beginning of this Section corresponds to β = 0 (the so called Landau gauge). In fact for β → 0 i 4 µ 2 − d x(∂µA ) µ lim e 2β → δ[∂µA ] (8.151) β→0 The choice corresponding to the quantization made in Section 1.5.3 is β =1.In this case 2 Gµν (x)=−gµν ∆F (x; m = 0) (8.152)

199 This is called the Feynman’s gauge. As we see the parameter β selects different covariant gauges. As already stressed the physics does not depend on β since the photon couples to the fermionic current ψγ¯ µψ, which is conserved in the abelian case, and as a consequence the β dependent part of the propagator, being proportional to kµkν does not contribute. The situation is quite different in the non-abelian case where the fermionic current is not conserved since the gauge fields are not neutral. Therefore we need further contributions in order to cancel the β dependent terms. Such contributions arise from the Faddeev-Popov determinant.

~~8.5 Path integral quantization of the non-abelian gauge theories~~

In the case of a non-abelian gauge theory there are several diﬀerences with respect to QED. The pure gauge lagrangian is not quadratic, it contains three-linear and four- linear interaction terms and furthermore the Faddeev-Popov determinant typically depends on the gauge ﬁelds. In this Section we will consider the pure gauge case, since the fermionic coupling is completely analogue to the abelian case. Separating the lagrangian in the quadratic part plus the rest, we get 1 µν LA = − FµνC FC 4 C 1 AB µ ν ν µ DE µ ν = − ∂µAνC − ∂νAµC − gf AµAAνB ∂ A − ∂ A − gf A A 4 C C C C D E (2) AB µ ν 1 2 AB DE µ ν = L + gf AµAAνB∂ A − g f f A A AµDAνE A C C 4 C C A B (2) I ≡LA + LA (8.153)

(2) (I) where LA and LA are the quadratic and the interacting part respectively. The interaction terms can be extracted by the functional integration by the usual trick (I) 1 δ iSA A i δηµ A Z[ηµ]=e Z0[ηµ ] (8.154) (I) 4 (I) with SA = d xLA and 1 µ 4 (2) 2 µ i d x LA − (∂µAA) 4 A i d xηAAµ 2β A Z0[ηµ]=N D(Aµ)e e ∆f [Aµ] (8.155) Consider now ∆f [Aµ]. In the Lorentz gauge

~~µ fA[Aµ]=∂µAA (8.156)~~

200 Under an inﬁnitesimal gauge transformation (for later convenience we have changed the deﬁnition of the gauge parameters, sending αA → gαA) µ AB µ µ δAC = gfC αAAB + ∂ αC (8.157) we have Ω µ AB µ fC [Aµ ]=∂µAC + gfC ∂µ(αAAB)+αC (8.158) Then Ω C δf [Aµ (x)] 2 4 BA 4 µ = δBCxδ (x − y)+gfC ∂µ(δ (x − y)AA)=MCB(x − y) (8.159) δαB(y) αA=0

We see that ∆f [A] depends explicitly on the gauge fields. In order to get an expression for the generating functional which is convenient fir deriving the Feynman rules, it is useful to give to it an exponential form. This can be done, recalling from eq. (7.76)) that a determinant can be written as a gaussian integral over Grassmann variables. ∗ 4 4 Ω i c (x)MAB(x − y)cB(y)d xd y δfA[Aµ (x)] ∗ A ∆f [Aµ]=det = N D(c, c )e δαB(y) αA=0 (8.160) The Grassmann fields cA(x) are called the ghost fields. It should be noticed that the ghost carry zero spin and therefore in their particle interpretation lead to a violation of the spin-statistics theorem. However this fact has no consequence, since our generating functional do not generate amplitudes with external ghosts. The action for the ghost fields can be read from the previous equation and we get 4 ∗ AB AC µ SFPG = d xcA(x) δ − gfB ∂µAC cB(x) (8.161)

The ghosts are zero mass particles interacting with the gauge fields through the interaction term (I) 4 ∗ AC µ SFPG = −g d xcAfB ∂µAC cB (8.162) In order to deal with this interaction it turns out convenient to define a new generating functional where we introduce also external sources for the ghost fields (I) 1 δ (I) −1 δ 1 δ iSA iSFPG , ∗ i δηµ i δη i δη ∗ Z[η, η∗,ηµ]=e e Z0[η, η ,ηµ] (8.163) with ∗ ∗ Z0[η, η ,ηµ]= D(Aµ)D(c, c ) 4 (2) 1 µ 2 A µ i d x L − (∂µA ) + η A A 2β A µ A ·e A 4 ∗ ∗ A A∗ i d x cAcA + cAη + η cA · e (8.164)

201 µ The integration over AA like in the abelian case. For the ghost ﬁeld we use eq. (7.76) with the result ∗ A i A A∗ 1 A ∗ ∗ A A∗ − (iη ) − (iη ) − ∗ i η η D(c, c )ei cA cA + cAη + η cA = e = e (8.165) Deﬁning − d4k e ikx ∆AB(x)=−δAB lim (8.166) →0+ (2π)4 k2 + i we get

i A µν ν − η (x)G (x − y)η (y) A∗ B ∗ µ AB B −i η (x)∆AB(x − y)η (y) Z0[η, η ,ηµ]=e 2 e (8.167) with µν µν GAB = δABG (x − y) (8.168) It is then easy to read the Feynman’s rules. For the propagators we get

~~. k~~

~~µ, A ν, B~~

~~Fig. 8.5.1 - Gauge ﬁeld propagator: kµkν 1 −iδAB gµν − (1 − β) (8.169) k2 + i k2 + i . . k~~

~~A B~~

Fig. 8.5.2 - Ghost ﬁeld propagator: 1 iδAB (8.170) k2 + i . The Feynman’s rules for the vertices are obtained by taking the Fourier transform AB of the interaction parts (here we have written fC ≡ fABC ).

~~202 . µ, A k 1~~

~~k3 λ, C~~

~~ν, B k 2~~

~~Fig. 8.5.3 - Three-linear gluon vertex:~~

~~4 4 gfABC(2π) δ (k1 + k2 + k3)[gµν (k1 − k2)λ + gνλ(k2 − k3)µ + gλµ(k3 − k1)ν] (8.171)~~

~~. µ, A ρ, D k 1 k 4~~

~~k3 ν, B λ, C k 2~~

~~Fig. 8.5.4 - Four-linear gluon vertex: 4 2 4 4 −ig (2π) δ ( ki) fABEfCDE(gµλgνρ − gνλgµρ) i=1 + fACEfBDE(gµν gλρ − gλνgµρ)+fADEfCBE(gµλgρν − gρλgµν ) (8.172)~~

~~. µ, A~~

~~p q~~

~~B C~~

~~Fig. 8.5.5 - Ghost-gluon vertex:~~

~~4 4 −gfABC (2π) δ (k + p − q)pµ (8.173)~~

203 When we have fermions coupled to the gauge ﬁelds we need to add the rule for the vertex (the propagator is the usual one). By a simple comparison with the coupling in QED we see that in this case the rule for the vertex is . µ, A

~~p p'~~

~~β,,b γ c~~

~~Fig. 8.5.6 - Fermion-gluon vertex:~~

~~A 4 4 −igTbc (γµ)βγ(2π) δ (p − p − k) (8.174)~~

~~8.6 The β-function in non-abelian gauge theories~~

We will skip a careful treatment of the renormalization in non-abelian gauge theories. However, since all the interaction terms have dimensions ≤ 4, the theory is renormalizable by power counting. On the other hand we will evaluate the beta function, since in these theories there is the possibility of having β<0, giving rise to asymptotic freedom. As it is known, QCD the theory of strong interactions is in fact a gauge theory based on the gauge group SU(3), and it is precisely ths feature of being an asymptotically free theory that has made it very attractive phenomenologically. In non-abelian gauge theories there are several ways to deﬁne the coupling, in fact we can look at the fermionic coupling, at the trilinear or at the four-linear couplings. If the theory is gauge invariant (and the renormalization should preserve this feature), all these ways should give the same result. This is in fact ensured by a set of identities among the renormalization constants (Ward identities), which are the analogue of the equality Z1 = Z2 in QED. Therefore we will study the deﬁni- tion coming from the fermionic coupling. In this case the relation between the bare coupling and the renormalized one is the same as in QED (see Section 6.3))

~~/2 Z1 B g = µ g 1/2 (8.175) Z2Z3 where Z1 is the vertex renormalization constant, and Z@ and Z3 are the wave function renormalization constants for the fermion and the gauge particle respectively.~~

204 As we shall see, one of the diﬀerences with QED is that now Z1 = Z2. As a consequence we will need to evaluate all the three renormalization constants. Let us start with the fermion self-energy. The one-loop contribution is given in Fig. 8.6.1. . k

~~p p p _ k Fig. 8.6.1 - One-loop fermion self-energy.~~

We will work in the Feynman-’t Hooft gauge, corresponding to the choice β =1for the gauge ﬁeld propagator (see eq. (8.169)). The Feynman rules for this diagram are the same as in QED except for the group factors. In order to avoid group theory complications, we will consider the case of a gauge group SU(2) with fermions in the spinor representation, where all the group factors are easily evaluated. At the same time we will exhibit also the general results for a theory SU(N)withfermionsin the fundamental representation (that is the one of dimensions N). Then, recalling the expression for the electron self-energy in QED (see eq. (2.166)), we get A A A A 1 − f Σ(p)= (T T )ΣQED(p)= (T T ) 2 (ˆp 4m)+ΣQED(p) (8.176) A A 8π

~~For SU(2) we have A A A A τ τ 3 (T T )ab = = δab (8.177) A A 2 2 ab 4~~

In the case of a group SU(N) and fermions in the fundamental representation it is possible to show that A A (T T )ab = C2(F )δab (8.178) A with N 2 − 1 C2(F )= (8.179) 2N A A This result is justiﬁed by the observation that the operator A T T (called the Casimir operator) commutes with all the generators and therefore it must be proportional to the identity matrix. From the previous expression for Σ we ﬁnd immediately (see Section 2.7) g2 Z =1− C (F ) (8.180) 2 8π2 2

~~205 .~~

~~= + +~~

~~++ + +~~

~~b) c) d)~~

~~+ +~~

~~e) f) Fig. 8.6.2 - One-loop expansion of the vacuum polarization diagram in non-abelian gauge theories.~~

The next task is the evaluation of Z3. To this end we need the one-loop expansion of the gauge particle propagator (the vacuum polarization diagram). This is given in Fig. 8.6.2. Since we are using dimensional regularization we can see immediately that the last three diagrams (diagrams d), e) and f)) of Fig. 8.6.2 are zero (this is not true in other regularizations). In fact all these diagrams involve the integral d2ωk (8.181) k2 which, in dimensional regularization gives (including a mass term) d2ωk = −iπωΓ(1 − ω)(a2)ω−1 (8.182) k2 − a2 By taking the limit a → 0, we see that for ω>1 the integral vanishes. Let us begin our calculation from the gauge particle contributions to the vacuum polarization. The kinematics is specified in Fig. 8.6.3. We will evaluate the truncated Green’s function which differs by a factor ig2 from the vacuum polarization vacuum Πµν as defined in Section 2.1. We have 2ω − − 2 (a) 1 2 ADC DBC d k i i ig Π = (−g) f f Eµν (8.183) µν,AB 2 (2π)2ω (p + k)2 k2

206 The factor 1/2 is a symmetry factor associated to the diagram and the tensor Eµν comes from the trilinear vertices. It is given by σ ρ σρ ρ σ Eµν = gµ(p − k) + g (2k + p)µ − gµ(2p + k) × − − [gσν(p k)ρ gνρ(2p + k)σ + gρσ(2k + p)ν ] 2 2 = gµν (p − k) +(2p + k) +2ω(2k + p)µ(2k + p)ν

~~− [(2p + k)µ(2k + p)ν − (2k + p)µ(p − k)ν +(p − k)µ(2p + k)ν ](8.184)~~

~~. p + k~~

~~C, ρ A, µ B, ν p D, σ p~~

~~k Fig. 8.6.3 - One-loop gauge particle contribution to the vacuum polarization.~~

ABC The group factor is easily evaluated for SU(2), since in this case f = ABC. Then we get ADCDBC = −2δAB (8.185) The general result for SU(N) is (taking into account that f ABC are completely antisymmetric) ADC DBC f f = −C2(G)δAB (8.186) with C2(G)=N (8.187) A ABC This result can be understood by noticing that (T )BC = if gives a representation of the Lie algebra of the group (the adjoint representation), and therefore C2(G) is nothing but the Casimir of the adjoint. By using the standard trick we reduce the two denominators in the integral to a single one 1 1 1 1 = dz (8.188) 2 2 2 − 2 2 k (k + p) 0 (k + z(1 z)p ) with k = k − pz (8.189)

Operating this substitution into the tensor Eµν , taking into account that the integration over the odd terms in k gives zero, and that for symmetry we can make the substitution gµν k k → k2 (8.190) µ ν 2ω

~~207 we get (by calling k2 with k2) 2 1 2 2 2 Eµν = gµν k 6 1 − + gµν p (1 + z) +(2− z) 2ω 2 2 + pµpν 2ω(1 − 2z) − 2 (1 + z) +(2− z)(1 − 2z) 2 ≡ aµν k + bµν (8.191)~~

Performing the integral over k with the formulas of Section 2.5 and putting everything together, we get ω 1 2 (a) 1 2 iπ 2 ig − g −C G δAB dz ωz − z p − ω aµν − ω bµν Πµν,AB = ( 2( )) 2ω (1 ) Γ(1 ) +Γ(2 ) 2 (2π) 0 (8.192) By taking only the 1/ term ( =4− 2ω) in the expansion and integrating over z we get 2 2 (a) g 19 µν 2 11 µ ν ig Π = i C (G)δAB g p − p p (8.193) µν,AB 16π2 2 6 3 As we see this contribution is not transverse, that is it is not proportional to 2 (pµpν − gµν p ), as it should be. in fact, we will see that adding to this diagram the contribution of the ghost loop we recover the transverse factor. The ghost contribution is the part b) of the expansion of Fig. 8.6.2. The relevant kinematics is given in Fig. 8.6.4. . p + k

~~D A, µ B, ν p p C~~

~~k Fig. 8.6.4 - One-loop ghost contribution to the vacuum polarization.~~

~~We have 2ω 2 (b) 2 ACD BDC d k i i ig Π =(−1)g f f (p + k)µkν (8.194) µν,AB (2π)2ω (p + k)2 k2~~

~~By the standard procedure we get (k = k − pz)) 1 ω 2 − µ ν − 2 (b) 2 d k kµkν p p z(1 z) ig Π = −g C (G)δAB dz (8.195) µν,AB 2 2ω 2 − 2 2 0 (2π) (k + z(1 z)p )~~

208 where we have eliminated the terms linear in k. Then, we perform the integral over k ω 1 2 (b) 2 π 1 2 − AB − − − µ ν − ig Πµν,AB = ig 2ω C2(G)δ dz z(1 z) p Γ(1 ω) p p Γ(2 ω) (2π) 0 2 (8.196) We perform now the integral in z. Then taking the limit ω → 2, and keeping only the leading term, we get 2 2 (b) g 1 2 1 ig Π = i C (G)δAB p gµν + pµpν (8.197) µν,AB 16π2 2 6 3 Putting together this contribution with the former one we get 2 2 (a) (b) g 5 2 ig (Π +Π )=−i C (G)δAB pµpν − gµν p (8.198) µν,AB µν,AB 8π2 3 2 The fermionic loop contribution is given in Fig. 8.6.5, and except for the group factor is the same evaluated in Section 2.6. . p + k

~~A, µ B, ν p p~~

~~k Fig. 8.6.5 - One-loop fermion contribution to the vacuum polarization.~~

The leading contribution is 2 2 (c) 2 A B QED g 2 ig Π = ig Tr(T T )Π = i nF δAB pµpν − gµν p (8.199) µν,AB µν 12π2 where we have used the conventional normalization for the group generators in the fermionic representation A B 1 Tr(T T )= nF δAB (8.200) 2 where nF is the number of fundamental representations. For instance, in the case of QCD, we have three different kind of quarks corresponding to different flavors. Then, the one-loop leading term contribution to the vacuum polarization is given by

~~2 2 (a) (b) (c) ig Πµν,AB = ig (Π +Π +Π ) µν,AB µν,AB µν,AB 2 g 5 2 2 = −i C (G) − nF pµpν − gµν p (8.201) 8π2 3 2 3~~

~~209 By comparison with Section 2.7 we get g2 5 2 Z =1+ C (G) − nF (8.202) 3 8π2 3 2 3~~

~~We are now left with the vertex corrections. The fermion contribution is given by the diagram of Fig. 8.6.6. .~~

~~A p' + k p + k B B~~

~~p' k p Fig. 8.6.6 - Fermion one-loop vertex correction.~~

~~Again, this is the same as in QED except for the group factors. the leading contribution is 3 µ,A g −ig3Λ = −i (T BT AT B)γµ (8.203) f 8π2 The group factor can be elaborated as follows~~

~~B A B B B A ABC C A ABC B C T T T = T (T T + if T )=C2(F)T + if T T 1 1 = C (F )T A − f ABCf BCDT D = C (F ) − C (G) T A(8.204) 2 2 2 2 2~~

~~Therefore we obtain 3 µ,A g 1 −ig3Λ = −i C (F ) − C (G) T Aγµ (8.205) f 8π2 2 2 2~~

The ﬁnal diagram to be evaluated is the vertex correction from the trilinear gauge coupling given in Fig. 8.6.7. We get (q = p − p) 2ω ˆ − 3 µ,A B d k ik i −ig Λ =(−igγνT ) g (2π)2ω k2 (p − k)2 ABC µν ρ νρ µ ρµ · (−g)f [g (q + p − k) − g (p + p − 2k) + g (p − k − q)ν] −i C · (−igγρT ) (8.206) (p − k)2

~~210 . A,µ~~

~~p' - k p - k~~

~~B C k p' p Fig. 8.6.7 - One-loop vertex correction from the trilinear gauge coupling.~~

Since we are interested only in the divergent piece, the calculation can be simpliﬁed a lot by the observation that the denominator goes as (k2)3, whereas the numerator goes at most as k2. Therefore the divergent piece is obtained by neglecting all the external momenta. Therefore the expression simpliﬁes to 2ω µν ρ νρ µ µρ ν d k γνkˆ(g k − 2g k + g k )γρ −ig3Λµ,A = ig3f ABC T BT C (8.207) g (2π)2ω (k2)3 The group factor gives i i f ABCT BT C = f ABCf BCDT D = C (G)T A (8.208) 2 2 2 whereas the numerator gives simply −3k2γµ.Thenwehave g3 3 d2ωk g3 3 −ig3Λµ,A = − C (G)T Aγµ = −i C (G)T Aγµ (8.209) g (2π)2ω 2 2 (k2)2 8π2 2 2 Adding up the two vertex contributions we obtain

3 µ,A g −ig3Λµ,A = −ig3(Λ +Λµ,A)=−i (C (F )+C (G)) T Aγµ (8.210) f g 8π2 2 2 Again, comparison with Section 2.7 gives g2 Z =1− (C (F )+C (G)) (8.211) 1 8π2 2 2 Using eq. (8.175) we have /2 1 /2 a1 gB = µ g(1 + ∆Z − ∆Z − ∆Z )=µ g + (8.212) 1 2 2 3 and collecting everything together we get g3 11 2 a = − C (G) − nF (8.213) 1 16π2 3 2 3

211 The evaluation of β(g) goes as in the QED case (see Section 6.3), that is ∂g(µ) 1 d 1 β(g)=µ = − 1 − g a = − (a − 3a )=a (8.214) ∂µ 2 dg 1 2 1 1 1 that is g3 11 2 β(g)=− C (G) − nF (8.215) 16π2 3 2 3 In the case of QCD where the gauge group is SU(3) we have

~~C2(G) = 3 (8.216) and therefore g3 2 β (g)=− 11 − nF (8.217) QCD 16π2 3 We see that QCD is asymptotically free (β(g) < 0) for 33 nF < (8.218) 2~~

~~212 Chapter 9~~

~~Spontaneous symmetry breaking~~

~~9.1 The linear σ-model~~

In this Section and in the following we will study, from a classical point of view, some field theory with particular symmetry properties. We will start examining the linear σ-model.ThisisamodelforN scalar fields, with a symmetry O(N). The lagrangian is given by N N N 2 1 µ 1 2 λ L = ∂µφi∂ φi − µ φiφi − φiφi (9.1) 2 2 4 i=1 i=1 i=1 This lagrangian is invariant under linear transformations acting upon the vector φ =(φ1, ···,φN ) and leaving invariant its norm N 2 |φ| = φiφi (9.2) i=1 Consider an infinitesimal transformation (from now on we will omit the index of sum over the indices which are repeated)

~~δφi = ijφj (9.3)~~

~~The condition for letting the norm invariant gives~~

~~|φ + δφ|2 = |φ|2 (9.4) from which φ · δφ =0 (9.5) or, in components, φiijφj =0 (9.6) This is satisﬁed by ij = −ji (9.7)~~

213 showing that the rotations in N dimensions depend on N(N −1)/2 parameters. For a ﬁnite transformation we have |φ|2 = |φ|2 (9.8) with φi = Sijφj (9.9) implying SST = 1 (9.10) In fact, by exponentiating the inﬁnitesimal transformation one gets

S = e (9.11) with T = − (9.12) implying that S is an orthogonal transformation. The matrices S form the rotation group in N dimensions, O(N). The Noether’s theorem implies a conserved current for any symmetry of the theory. In this case we will get N(N − 1)/2 conserved quantities. It is useful for further generalizations to write the inﬁnitesimal transformation in the form

i AB δφi = ijφj = − ABT φj,i,j=1, ···,N, A,B =1, ···,N (9.13) 2 ij which is similar to what we did in Section 3.3 when we discussed the Lorentz transformations. By comparison we see that the matrices T AB are given by

~~AB A B A B Tij = i(δi δj − δj δi ) (9.14)~~

~~It is not diﬃcult to show that these matrices satisfy the algebra~~

~~[T AB,TCD]=−iδAC T BD + iδADT BC − iδBDT AC + iδBCT AD (9.15)~~

This is nothing but the Lie algebra of the group O(N), and the T AB are the infinitesimal generators of the group. Applying now the Noether’s theorem we find the conserved current L ∂ − i AB jµ = µ δφi = φi,µABTij φj (9.16) ∂(∂ φi) 2 since the N(N − 1)/2 parameters AB are linearly independent, we find the N(N − 1)/2 conserved currents AB AB Jµ = −iφi,µTij φj (9.17) InthecaseofN = 2 the symmetry is the same as for the charged field in a real basis, and the only conserved current is given by

~~12 21 Jµ = −Jµ = φ1,µφ2 − φ2,µφ1 (9.18)~~

214 One can easily check that the charges associated to the conserved currents close the same Lie algebra as the generators T AB. More generally, if we have conserved currents given by A A jµ = −iφi,µTij φj (9.19) with [T A,TB]=if ABCT C (9.20) then, using the canonical commutation relations, we get

~~[QA,QB]=if ABC QC (9.21) with A 3 A − 3 ˙ A Q = d xj0 (x)= i d x φiTij φj (9.22)~~

~~A particular example is the case N = 4. We parameterize our ﬁelds in the form φ =(π1,π2,π3,σ)=(π, σ) (9.23)~~

~~These ﬁelds can be arranged into a 2 × 2 matrix~~

~~M = σ + iτ · π (9.24) where τ are the Pauli matrices. Noticing that τ2 is pure imaginary, τ1 and τ3 real, and that τ2 anticommutes with τ1 and τ3,weget~~

~~∗ M = τ2M τ2 (9.25)~~

Furthermore we have the relation 1 |φ|2 = σ2 + |π|2 = Tr(M †M) (9.26) 2 Using this it is easy to write the lagrangian for the σ-model in the form 1 † µ 1 2 † 1 † 2 L = Tr(∂µM ∂ M) − µ Tr(M M) − λ Tr(M M) (9.27) 4 4 16 This lagrangian is invariant under the following transformation of the matrix M

M → LMR† (9.28) where L and R are two special (that is with determinant equal to 1) unitary matrices, that is L, R ∈ SU(2). The reason to restrict these matrices to be special is that only in this way the transformed matrix satisfy the condition (9.25). In fact, if A is a2× 2 matrix with detA =1,then

~~T −1 τ2A τ2 = A (9.29)~~

~~215 Therefore, for M = LMR† we get~~

~~∗ ∗ ∗ T ∗ ∗ T τ2M τ2 = τ2L M R τ2 = τ2L τ2(τ2M τ2)τ2R τ2 (9.30) and from (9.29)~~

∗ †T †−1 τ2L τ2 = τ2L τ2 = L = L T −1 † τ2R τ2 = R = R (9.31) since the L and R are independent transformations, the invariance group in this basis is SU(2)L ⊗ SU(2)R. In fact this group and O(4) are related by the following observation: the transformation M → LMR† is a linear transformation on the matrix elements of M, but from the relation (9.26) we see that M → LMR† leaves the norm of the vector φ =(π, σ) invariant and therefore the same must be true for the linear transformation acting upon the matrix elements of M,thatisonσ and π. Therefore this transformation must belongs to O(4). This shows that the two groups SU(2) ⊗ SU(2) and O(4) are homomorphic (actually there is a 2 to 1 relationship, since −L and −R define the same S as L and R). We can evaluate the effect of an infinitesimal transformation. To this end we will consider separately left and right transformations. We parameterize the transformations as follows i i L ≈ 1 − θL · τ, R ≈ 1 − θR · τ (9.32) 2 2 then we get i i 1 i δLM =(− θL · τ)M =(− θL · τ)(σ + iπ · τ)= θL · π + (θL ∧ π − θLσ) · τ (9.33) 2 2 2 2 wherewehaveused τiτj = δij + iijkτk (9.34) Since δLM = δLσ + iδLπ · τ (9.35) we get 1 1 δLσ = θL · π, δLπ = (θL ∧ π − θLσ) (9.36) 2 2 Analogously we obtain 1 1 δRσ = − θR · π, δRπ = (θR ∧ π + θRσ) (9.37) 2 2 The combined transformation is given by 1 1 δσ = (θL − θR) · π, δπ = [(θL + θR) ∧ π − (θL − θR)σ] (9.38) 2 2

216 and we can check immediately that σδσ + π · δπ = 0 (9.39) as it must be for a transformation leaving the form σ2 +|π|2 invariant. Of particular interest are the transformations with θL = θR ≡ θ.InthiscasewehaveL = R and M → LML† (9.40)

These transformations span a subgroup SU(2) of SU(2)L ⊗ SU(2)R called the diagonal subgroup. In this case we have δσ =0,δπ = θ ∧ π (9.41) We see that the transformations corresponding to the diagonal SU(2) are the rotations in the 3-dimensional space spanned by π. These rotations deﬁne a subgroup O(3) of the original symmetry group O(4). From the Noether’s theorem we get the conserved currents

L 1 1 j = σ,µθL · π + π,µ · (θL ∧ π − θLσ) (9.42) µ 2 2 and dividing by θL/2 L Jµ = σ,µπ − π,µσ − π,µ ∧ π (9.43) and analogously R Jµ = −σ,µπ + π,µσ − π,µ ∧ π (9.44) Using the canonical commutation relations one can verify that the corresponding charges satisfy the Lie algebra of SU(2)L ⊗ SU(2)R

L L L R R R L R [Qi ,Qj ]=iijkQk , [Qi ,Qj ]=iijkQk , [Qi ,Qj ] = 0 (9.45) By taking the following combinations of the currents 1 1 JV = (JL + JR), JA = (JL − JR) (9.46) µ 2 µ µ µ 2 µ µ one has V Jµ = π ∧ π,µ (9.47) and A Jµ = σ,µπ − π,µσ (9.48) The corresponding algebra of charges is

~~V V V V A A A A V [Qi ,Qj ]=iijkQk , [Qi ,Qj ]=iijkQk , [Qi ,Qj ]=iijkQk (9.49)~~

~~V These equations show that Qi are the inﬁnitesimal generators of a subgroup SU(2) of SU(2)L ⊗ SU(2)R which is the diagonal subgroup, as it follows from~~

~~V V [Qi ,πj]=iijkπk, [Qi ,σ] = 0 (9.50)~~

217 In the following we will be interested in treating the interacting field theories by using the perturbation theory. As in the quantum mechanical case, this is well defined only when we are considering the theory close to a minimum the energy of the system. In fact if we are going to expand around a maximum the oscillation of the system can become very large leading us outside of the domain of perturbation theory. In the case of the linear σ-model the energy is given by N L N 3 3 ∂ 3 1 2 2 2 H = d x H = d x φ˙i −L = d x (φ˙ + |∇ φi| )+V (|φ| ) ˙ 2 i i=1 ∂φi i=1 (9.51) Since in the last member of this relation the first two terms are positive definite, it follows that the absolute minimum is obtained for constant field configurations, such that ∂V (|φ|2) = 0 (9.52) ∂φi

Let us call by vi the generic solution to this equation (in general it could happen that the absolute minimum is degenerate). Then the condition for getting a minimum is that the eigenvalues of the matrix of the second derivatives of the potential at the stationary point are definite positive. In this case we define new fields by shifting the original fields by φi → φi = φi − vi (9.53) The lagrangian density becomes

~~1 µ 2 L = ∂µφ ∂ φ − V (|φ + v| ) (9.54) 2 i i Expanding V in series of φi we get~~

~~2 2 1 ∂ V V = V (|v| )+ φiφj + ··· (9.55) 2 ∂φi∂φj φ=v~~

This equation shows that the particle masses are given by the eigenvalues of the second derivative of the potential at the minimum. In the case of the linear σ-model we have 1 λ V = µ2|φ|2 + (|φ|2)2 (9.56) 2 4 Therefore ∂V 2 2 = µ φi + λφi|φ| (9.57) ∂φi

~~In order to have a solution to the stationary condition we must have φi =0,or~~

~~µ2 |φ|2 = − (9.58) λ~~

218 This equation has real solutions only if µ2/λ < 0. However, in order to have a potential bounded from below one has to require λ>0, therefore we may have non zero solutions to the minimum condition only if µ2 < 0. But, notice that in this case, µ2 cannot be identified with a physical mass, these are given by the eigenvalues of the matrix of the second derivatives of the potential at the minimum and they are positive definite by definition. We will study this case in the following Sections. 2 Inthecaseofµ > 0 the minimum is given by φi = 0 and one can study the theory by taking the term λ(|φ|2)2 as a small perturbation (that is requiring that both λ and the values of φi, the fluctuations, are small). The free theory is given by the quadratic terms in the lagrangian density, and they describe N particles of common mass m. Furthermore, both the free and the interacting theories are O(N) symmetric.

~~9.2 Spontaneous symmetry breaking~~

In this Section we will see that the linear σ-model with µ2 < 0, is just but an example of a general phenomenon which goes under the name of spontaneous symmetry breaking of symmetry. This phenomenon lies at the basis of the modern description of phase transitions and it has acquired a capital relevance in the last years in all ﬁeld of physics. The idea is very simple and consists in the observation that a theory with hamiltonian invariant under a symmetry group may not show explicitly the symmetry at the level of the solutions. As we shall see this may happen when the following conditions are realized:

~~• The theory is invariant under a symmetry group G.~~

~~• The fundamental state of the theory is degenerate and transforms in a non trivial way under the symmetry group.~~

~~Just as an example consider a scalar ﬁeld described by a lagrangian invariant under parity P : φ →−φ (9.59) The lagrangian density will be of the type~~

~~1 µ 2 L = ∂µφ∂ φ − V (φ ) (9.60) 2 If the vacuum state is non degenerate, barring a phase factor, we must have~~

~~P |0 = |0 (9.61) since P commutes with the hamiltonian. It follows~~

~~0|φ|0 = 0|P −1PφP−1P |0 = 0|PφP−1|0 = − 0|φ|0 (9.62)~~

219 from which 0|φ|0 = 0 (9.63) Things change if the fundamental state is degenerate. This would be the case in the example (9.60), if µ2 λ V (φ2)= φ2 + φ4 (9.64) 2 4 with µ2 < 0. In fact, this potential has two minima located at µ2 φ = ±v, v = − (9.65) λ By denoting with |R e |L the two states corresponding to the classical configurations φ = ±v,wehave P |R = |L = |R (9.66) Therefore R|φ|R = R|P −1PφP−1P |R = − L|φ|L (9.67) which now does not imply that the expectation value of the field vanishes. In the following we will be rather interested in the case of continuous symmetries. So let us consider two scalar fields, and a lagrangian density with symmetry O(2)

1 µ 1 2 λ 2 L = ∂µφ · ∂ φ − µ φ · φ − (φ · φ) (9.68) 2 2 4 where · 2 2 φ φ = φ1 + φ2 (9.69) For µ2 > 0 there is a unique fundamental state (minimum of the potential) φ =0, whereas for µ2 < 0 there are infinite degenerate states given by ||2 2 2 2 φ = φ1 + φ2 = v (9.70) with v defined as in (9.65). By denoting with R(θ) the operator rotating the fields in the plane (φ1,φ2), in the non-degenerate case we have R(θ)|0 = |0 (9.71) and 0|φ|0 = 0|R−1RφR−1R|0 = 0|φθ|0 = 0 (9.72) since φθ = φ.Inthecaseµ2 < 0 (degenerate case), we have

~~R(θ)|0 = |θ (9.73) where |θ is one of the inﬁnitely many degenerate fundamental states lying on the circle |φ|2 = v2.Then~~

~~−1 −1 θ 0|φi|0 = 0|R (θ)R(θ)φiR (θ)R(θ)|0 = θ|φi |θ (9.74)~~

220 with θ −1 φi = R(θ)φiR (θ) = φi (9.75) Again, the expectation value of the ﬁeld (contrarily to the non-degenerate state) does not need to vanish. The situation can be described qualitatively saying that the existence of a degenerate fundamental state forces the system to choose one of these equivalent states, and consequently to break the symmetry. But the breaking is only at the level of the solutions, the lagrangian and the equations of motion preserve the symmetry. One can easily construct classical systems exhibiting spontaneous symmetry breaking. For instance, a classical particle in a double-well potential. This system has parity invariance x →−x,wherex is the particle position. The equilibrium positions are around the minima positions, ±x0. If we put the particle close to x0, it will perform oscillations around that point and the original symmetry is lost. A further example is given by a ferromagnet which has an hamiltonian invariant under rotations, but below the Curie temperature exhibits spontaneous magnetization, breaking in this way the symmetry. These situations are typical for the so called second order phase transitions. One can describe them through the Landau free-energy, which depends on two diﬀerent kind of parameters:

~~• Control parameters,asµ2 for the scalar ﬁeld, and the temperature for the ferromagnet.~~

~~• Order parameters, as the expectation value of the scalar ﬁeld or as the magnetization.~~

The system goes from one phase to another varying the control parameters, and the phase transition is characterized by the order parameters which assume different values in different phases. In the previous examples, the order parameters were zero in the symmetric phase and different from zero in the broken phase. The situation is slightly more involved at the quantum level, since spontaneous symmetry breaking cannot happen in finite systems. This follows from the existence of the tunnel effect. Let us consider again a particle in a double-well potential, and recall that we have defined the fundamental states through the correspondence with the classical minima

~~x = x0 →|R~~

~~x = −x0 →|L (9.76)~~

But the tunnel eﬀect gives rise to a transition between these two states and as a consequence it removes the degeneracy. In fact, due to the transition the hamiltonian acquires a non zero matrix element between the states |R and |L .Bydenoting with H the matrix of the hamiltonian between these two states, we get H = 0 1 (9.77) 1 0

221 The eigenvalues of H are (0 + 1,0 − 1) (9.78) We have no more degeneracy and the eigenstates are 1 |S = √ (|R + |L ) (9.79) 2 with eigenvalue ES = 0 + 1,and 1 |A = √ (|R −|L ) (9.80) 2 with eigenvalue EA = 0 − 1. One can show that 1 < 0 and therefore the fundamental state is the symmetric one, |S . This situation gives rise to the well known effect of quantum oscillations. We can express the states |R and |L in terms of the energy eigenstates |R = √1 (|S + |A ) 2 |L = √1 (|S −|A ) (9.81) 2 Let us now prepare a state, at t = 0, by putting the particle in the right minimum. This is not an energy eigenstate and its time evolution is given by 1 − − 1 − − |R, t = √ e iES t|S + e iEAt|A = √ e iES t |S + e it∆E|A (9.82) 2 2 with ∆E = EA − ES. Therefore, for t = π/∆E the state |R transforms into the state |L . The state oscillates with a period given by 2π T = (9.83) ∆E In nature there are finite systems as sugar molecules, which seem to exhibit spontaneous symmetry breaking. In fact one observes right-handed and left-handed sugar molecules. The explanation is simply that the energy difference ∆E is so small that the oscillation period is of the order of 104 − 106 years. The splitting of the fundamental states decreases with the height of the potential between two minima, therefore, for infinite systems, the previous mechanism does not work, and we may have spontaneous symmetry breaking. In fact, coming back to the scalar field example, its expectation value on the vacuum must be a constant, as it follows from the translational invariance of the vacuum − 0|φ(x)|0 = 0|eiP xφ(0)e iP x|0 = 0|φ(0)|0 = v (9.84) and the energy difference between the maximum at φ = 0, and the minimum at φ = v, becomes infinite in the limit of infinite volume µ2 λ µ4 µ4 H(φ =0)− H(φ = v)=− d3x v2 + v4 = d3x = V (9.85) V 2 4 4λ V 4λ

~~222 9.3 The Goldstone theorem~~

From our point of view, the most interesting consequence of spontaneous symmetry breaking is the Goldstone theorem. This theorem says that for any continuous symmetry spontaneously broken, there exists a massless particle (the Goldstone boson). The theorem holds rigorously in a local ﬁeld theory, under the following hypotheses

~~• The spontaneous broken symmetry must be a continuous one.~~

~~• The theory must be manifestly covariant.~~

~~• The Hilbert space of the theory must have a deﬁnite positive norm.~~

We will limit ourselves to analyze the theorem in the case of a classical scalar ﬁeld theory. Let us start considering the lagrangian for the linear σ-model with invariance O(N) 2 1 µ µ λ 2 L = ∂µφ · ∂ φ − φ · φ − (φ · φ) (9.86) 2 2 4 The conditions that V must satisfy in order to have a minimum are

∂V 2 2 = µ φl + λφl|φ| = 0 (9.87) ∂φl with solutions − 2 2 2 µ φl =0, |φ| = v ,v= (9.88) λ The character of the stationary points can be studied by evaluating the second derivatives 2 ∂ V 2 2 = δlm(µ + λ|φ| )+2λφlφm (9.89) ∂φl∂φm We have two possibilities

~~• µ2 > 0, we have only one real solution given by φ = 0, which is a minimum, since 2 ∂ V 2 = δlmµ > 0 (9.90) ∂φl∂φm~~

• µ2 < 0, there are inﬁnite solutions, among which φ = 0 is a maximum. The points of the sphere |φ|2 = v2 are degenerate minima. In fact, by choosing φl = vδlN as a representative point, we get

~~2 ∂ V 2 =2λv δlN δmN > 0 (9.91) ∂φl∂φm~~

223 Expanding the potential around this minimum we get 1 ∂2V V (φ) ≈ V + (φl − vδlN )(φm − vδmN ) (9.92) minimum 2 ∂φl∂φm minimum If we are going to make a perturbative expansion, the right ﬁelds to be used are φl − vδlN , and their mass is just given by the coeﬃcient of the quadratic term ⎡ ⎤ · 00 0 2 ⎢ · ⎥ 2 ∂ V 2 ⎢ 00 0 ⎥ Mlm = = −2µ δlN δmN = ⎣ ⎦ (9.93) ∂φl∂φm minimo ··· · 00·−2µ2

Therefore the masses of the fields φa, a =1, ···,N − 1,andχ = φN − v,aregiven by 2 2 − 2 mφa =0,mχ = 2µ (9.94) By defining m2 = −2µ2 (9.95) we can write the potential as a function of the new fields N− N− 2 m4 1 m2λ 1 λ 1 V = − + m2χ2 + χ φ2 + χ2 + φ2 + χ2 (9.96) 16λ 2 2 a 4 a a=1 a=1

In this form the original symmetry O(N) is broken. However a residual symmetry − N−1 2 O(N 1) is left. In fact, V depends only on the combination a=1 φa, and it is invariant under rotations around the axis we have chosen as representative for the fundamental state, (0, ···,v). It must be stressed that this is not the most general potential invariant under O(N − 1). In fact the most general potential (up to the fourth order in the fields) describing N scalar fields with a symmetry O(N − 1) would depend on 7 coupling constants, whereas the one we got depends only on the two parameters m and λ. Therefore spontaneous symmetry breaking puts heavy constraints on the dynamics of the system. We have also seen that we have N − 1 massless scalars. Clearly the rotations along the first N − 1 directions leave the potential invariant, whereas the N − 1 rotations on the planes a − N move away from the surface of the minima. This can be seen also in terms of generators. Since the field we have chosen as representative of the ground state is φi|min = vδiN ,we have ab a b b a Tij φj|min = i(δi δj − δi δj )vδjN = 0 (9.97) since a, b =1, ···,N − 1, and

~~aN a N N a a Tij φj|min = i(δi δj − δi δj )vδjN = ivδi = 0 (9.98)~~

~~Therefore we have N − 1 broken symmetries and N − 1 massless scalars. The generators of O(N) divide up naturally in the generators of the vacuum symmetry~~

224 (here O(N −1)), and in the so called broken generators, each of them corresponding to a massless Goldstone boson. In general, if the original symmetry group G of the theory is spontaneously broken down to a subgroup H (which is the symmetry of the vacuum), the Goldstone bosons correspond to the generators of G which are left after subtracting the generators of H. Intuitively one can understand the origin of the massless particles noticing that the broken generators allow transitions from a possible vacuum to another. Since these states are degenerate the operation does not cost any energy. From the relativistic dispersion relation this implies that we must have massless particles. One can say that Goldstone bosons correspond to ﬂat directions in the potential.

~~9.4 The Higgs mechanism~~

We have seen that the spontaneous symmetry breaking mechanism, in the case of continuous symmetry leads to massless scalar particles, the Goldstone bosons. Also gauge theories lead to massless vector bosons, in fact, as in the electromagnetic case, gauge invariance forbids the presence in the lagrangian of terms quadratic in the fields. Unfortunately in nature the only massless particles we know are the photon and perhaps the neutrinos, which however are fermions. But once one couples spontaneous symmetry breaking to a gauge symmetry, things change. In fact, if we look back at the hypotheses underlying a gauge theory, it turns out that Goldstone theorem does not hold in this context. The reason is that it is impossible to quantize a gauge theory in a way which is at the same time manifestly covariant and has a Hilbert space with positive definite metric. This is well known already for the electromagnetic field, where one has to choose the gauge before quantization. What happens is that, if one chooses a physical gauge, as the Coulomb gauge, in order to have a Hilbert space spanned by only the physical states, than the theory looses the manifest covariance. If one goes to a covariant gauge, as the Lorentz one, the theory is covariant but one has to work with a big Hilbert space, with non-definite positive metric, and where the physical states are extracted through a supplementary condition. The way in which the Goldstone theorem is evaded is that the Goldstone bosons disappear, and, at the same time, the gauge bosons corresponding to the broken symmetries acquire mass. This is the famous Higgs mechanism. Let us start with a scalar theory invariant under O(2)

2 1 µ µ λ 2 L = ∂µφ · ∂ φ − φ · φ − (φ · φ) (9.99) 2 2 4 and let us analyze the spontaneous symmetry breaking mechanism. If µ2 < 0the symmetry is broken and we can choose the vacuum as the state −µ2 φ =(v, 0),v= (9.100) λ

225 2 2 After the translation φ1 = χ+v,with 0|χ|0 =0,wegetthepotential(m = −2µ ) m4 1 m2λ λ V = − + m2χ2 + χ(φ2 + χ2)+ (φ2 + χ2)2 (9.101) 16λ 2 2 2 4 2 In this case the group O(2) is completely broken (except for the discrete symmetry φ2 →−φ2). The Goldstone ﬁeld is φ2 . This has a peculiar way of transforming under O(2). In fact, the original ﬁelds transform as

δφ1 = −αφ2,δφ2 = αφ1 (9.102) from which δχ = −αφ2,δφ2 = αχ + αv (9.103) We see that the Goldstone field undergoes a rotation plus a translation, αv.This is the main reason for the Goldstone particle to be massless. In fact one can have invariance under translations of the field, only if the potential is flat in the corresponding direction. This is what happens when one moves in a way which is tangent to the surface of the degenerate vacuums (in this case a circle). How do things change if our theory is gauge invariant? In that case we should have invariance under a transformation of the Goldstone field given by

~~δφ2(x)=α(x)χ(x)+α(x)v (9.104)~~

Since α(x) is an arbitrary function of the space-time point, it follows that we can choose it in such a way to make φ2(x) vanish. In other words it must be possible to eliminate the Goldstone field from the theory. This is better seen by using polar coordinates for the fields, that is φ ρ = φ2 + φ2, sin θ = 2 (9.105) 1 2 2 2 φ1 + φ2 Under a finite rotation, the new fields transform as

~~ρ → ρ, θ → θ + α (9.106)~~

It should be also noticed that the two coordinate systems coincide when we are close to the vacuum, as when we are doing perturbation theory. In fact, in that case we can perform the following expansion φ φ ρ = φ2 + χ2 +2χv + v2 ≈ v + χ, θ ≈ 2 ≈ 2 (9.107) 2 v + χ v Again, if we make the theory invariant under a local transformation, we will have invariance under θ(x) → θ(x)+α(x) (9.108)

226 By choosing α(x)=−θ(x) we can eliminate this last ﬁeld from the theory. The only remaining degree of freedom in the scalar sector is ρ(x). Let us study the gauging of this model. It is convenient to introduce complex variables 1 † 1 φ = √ (φ1 + iφ2),φ= √ (φ1 − iφ2) (9.109) 2 2 The O(2) transformations become phase transformations on φ

~~φ → eiαφ (9.110) and the lagrangian (9.99) can be written as~~

~~† µ 2 † † 2 L = ∂µφ ∂ φ − µ φ φ − λ(φ φ) (9.111)~~

~~We know that it is possible to promote a global symmetry to a local one by introducing the covariant derivative~~

~~∂µφ → (∂µ − igAµ)φ (9.112) from which~~

~~† µ µ 2 † † 2 1 µν L =(∂µ + igAµ)φ (∂ − igA )φ − µ φ φ − λ(φ φ) − Fµν F (9.113) 4 In terms of the polar coordinates (ρ, θ)wehave~~

~~1 1 − φ = √ ρeiθ,φ† = √ ρe iθ (9.114) 2 2 By performing the following gauge transformation on the scalars~~

~~− φ → φ = φe iθ (9.115) and the corresponding transformation on the gauge ﬁelds~~

~~1 Aµ → A = Aµ − ∂µθ (9.116) µ g~~

~~the lagrangian will depend only on the ﬁelds ρ and Aµ (we will put again Aµ = Aµ)~~

~~2 1 µ µ 2 λ 4 1 µν L = (∂µ + igAµ)ρ(∂ − igAµ)ρ − ρ − ρ − Fµν F (9.117) 2 2 4 4 In this way the Goldstone boson disappears. We have now to translate the ﬁeld ρ~~

~~ρ = χ + v, 0|χ|0 = 0 (9.118)~~

~~227 and we see that this generates a bilinear term in Aµ, coming from the covariant derivative, given by 1 2 2 µ g v AµA (9.119) 2 Therefore the gauge ﬁeld acquires a mass~~

2 2 2 mA = g v (9.120) It is instructive to count the degrees of freedom before and after the gauge transformation. Before we had 4 degrees of freedom, two from the scalar fields and two from the gauge field. After the gauge transformation we have only one degree of freedom from the scalar sector, but three degrees of freedom from the gauge vector, because now it is a massive vector field. The result looks a little bit strange, but the reason why we may read clearly the number of degrees of freedom only after the gauge transformation is that before the lagrangian contains a mixing term

µ Aµ∂ θ (9.121) between the Goldstone ﬁeld and the gauge vector which makes complicate to read the mass of the states. The previous gauge transformation realizes the purpose of making that term vanish. The gauge in which such a thing happens is called the unitary gauge. We will consider now the further example of a symmetry O(N). The lagrangian invariant under local transformations is 2 1 µ µ λ 2 L = (Dµ)ijφj(D )ikφk − φiφi − (φiφi) (9.122) 2 2 4 where g AB AB (Dµ)ij = δij∂µ + i (T )ijW (9.123) 2 µ AB A B A B 2 where (T )lm = i(δl δm − δmδl ). In the case of broken symmetry (µ < 0), we choose again the vacuum along the direction N,withv deﬁned as in (9.100)

~~φi = vδiN (9.124)~~

Recalling that ab aN a Tij φj =0,Tij φj = ivδi ,a,b=1, ···,N − 1 (9.125) min min the mass term for the gauge ﬁeld is given by 1 2 AB CD AB µCD − g Tij φj (T )ikφk Wµ W 8 min min 1 2 aN bN aN µbN = − g Tij φj (T )ikφk Wµ W 2 min min 1 1 = g2v2δaδbW aN W µbN = g2v2W aN W µaN (9.126) 2 i i µ 2 µ

228 aN aN Therefore, the fields Wµ associated to the broken directions T acquire a mass 2 2 ab g v /2, whereas Wµ , associated to the unbroken symmetry O(N −1), remain massless. In general, if G is the global symmetry group of the lagrangian, H the subgroup of G leaving invariant the vacuum, and GW the group of local (gauge) symmetries, GW ∈ G, one can divide up the broken generators in two categories. In the first category fall the broken generators lying in GW ; they have associated massive vector bosons. In the second category fall the other broken generators; they have associated massless Goldstone bosons. Finally the gauge fields associated to generators of GW lying in H remain massless. From the previous derivation this follows noticing that the generators of H annihilate the minimum of the fields, leaving the corresponding gauge bosons massless, whereas the non zero action of the broken generators generate a mass term for the other gauge fields. The situation is represented in Fig. 9.4.1.

~~H G W~~

Fig. 6.1 -This ﬁgure shows the various groups, G, the global symmetry of the lagrangian, H ∈ G,thesymmetryofthevacuum,andGW , the group of local symmetries. The broken generators in GW correspond to massive vector bosons. The broken generators do not belonging to GW correspond to massless Goldstone bosons. Th unbroken generators in GW correspond to massless vector bosons.

We can now show how to eliminate the Goldstone bosons. In fact we can define new fields ξa and χ as aN −iT ξa φi = e (χ + v) (9.127) iN where a =1, ···,N − 1, that is the sum is restricted to the broken directions. The other degree of freedom is in the other factor. The correspondence among the fields

229 φ and (ξa,χ) can be seen easily by expanding around the vacuum aN −iT ξa aN a e ≈ δiN − i(T )iN ξa = δiN + δi ξa (9.128) iN from which φi ≈ (vξa,χ+ v) (9.129) showing that the ξa’s are really the Goldstone fields. The unitary gauge is defined through the transformation aN iT ξa φi → e φj = δiN (χ + v) (9.130) ij aN aN aN aN iT ξa −iT ξa i iT ξa −iT ξa Wµ → e Wµe − ∂µe e (9.131) g This transformation eliminates the Goldstone degrees of freedom and the resulting aN lagrangian depends on the field χ, on the massive vector fields Wµ and on the ab massless field Wµ . Notice again the counting of the degrees of freedom N +2N(N − 1)/2=N 2 in a generic gauge, and 1 + 3(N − 1) + 2(N − 1)(N − 2)/2=N 2 in the unitary gauge.

9.5 Quantization of a spontaneously broken gauge theory in the Rξ-gauge In the previous Section we have seen that in the unitary gauge some of the gauge fields acquire a mass. The corresponding quadratic part of the lagrangian will be (we consider here the abelian case corresponding to a symmetry U(1) or O(2)) 2 1 µν m 2 L = − Fµν F + W (9.132) 4 2 where Wµ is the massive gauge field (called also a Yang-Mills massive field). The free equations of motion are

~~2 ν ( + m )Wµ − ∂µ(∂ Wν) = 0 (9.133) Let us study the solutions of this equation in momentum space. We deﬁne 4 −ikx Wµ(x)= d kWµ(k)e (9.134)~~

~~It is convenient to introduce a set of four independent four-vectors. By putting kµ =(E,k) we introduce~~

~~µ(i) =(0,ni),i=1 , 2 1 k µ(3) = |k|,E (9.135) m |k|~~

~~230 with k · ni =0,ni · nj = δij (9.136)~~

~~Then we can decompose Wµ(k) as follows~~

~~µ(λ) Wµ(k)= aλ(k)+kµb(k) (9.137)~~

~~µ(λ) Since kµ = 0, from the equation of motion we get~~

2 2 µ(λ) 2 (−k + m )( aλ(k)+kµb(k)) + kµk b(k) = 0 (9.138) implying 2 2 2 (k − m )aλ(k)=0,mb(k) = 0 (9.139) Therefore, for m = 0 the ﬁeld has three degrees of freedom with k2 = m2, corresponding to transverse and longitudinal polarization. When the momentum is on shell, we can deﬁne a fourth unit vector kµ µ(0) = (9.140) m In this way we have four independent unit vectors which satisfy the completeness relation 3 µ(λ) ν(λ) µν gλλ = g (9.141) λ,λ=0 Therefore the sum over the physical degrees of freedom gives

~~3 kµkν µ(λ)ν(λ) = −gµν + (9.142) m2 λ=1 The propagator is deﬁned by the equation 2 νλ λ 4 ( + m ) − ∂µ∂ν G (x)=igµδ (x) (9.143)~~

~~Solving in momentum space we ﬁnd 4 µ ν µν d k − i µν k k G (k)= e ikx −g + (9.144) (2π)4 k2 − m2 + i m2~~

Therefore, the corresponding Feynman’s rule for the massive gauge ﬁeld propagator is i kµkν − gµν − (9.145) k2 − m2 + i m2 Notice that the expression in parenthesis is nothing but the projector over the physical states given in eq. (9.142). This propagator has a very bad high-energy behaviour. In fact it goes to constant for k →∞. As a consequence the argument

231 for the renormalizability of the theory based on power counting does not hold any more. Although in the unitary gauge a spontaneously broken gauge theory does not seem to be renormalizable, we have to take into account that the gauge invariance allows us to make diﬀerent choices of the gauge. In fact we will show that there exist an entire class of gauges where the theory is renormalizable by power counting. We will discuss this point only for the abelian case, with gauge group O(2), but the argument cab be easily extended to the non-abelian case. Let us consider the O(2) theory as presented in Section 9.4. Let us recall that under a transformation of O(2) the two real scalar ﬁeld of the theory transform as

~~δφ1 = −αφ2,δφ2 = αφ1 (9.146) Therefore, in order to make it a gauge theory we introduce covariant derivatives~~

~~Dµφ1 = ∂µφ1 + gWµφ2~~

~~Dµφ2 = ∂µφ2 − gWµφ1 (9.147) with Wµ transforming under a local transformation as 1 δWµ(x)= ∂µα(x) (9.148) g~~

Assuming that the spontaneous breaking occurs along the direction φ1, we introduce the fields φ1 = v + χ, φ2 = φ (9.149) with µ2 v2 = − (9.150) λ Remember that χ is the Higgs field, whereas φ is the Goldstone field. In terms of the new variables the lagrangian is

1 2 1 2 1 µν L = (∂µχ + gWµφ) + (∂µφ − gWµχ − gvWµ) − V (χ, φ) − Fµν F (9.151) 2 2 4 where, the potential up to quadratic terms is given by (see eq. (9.101)) m4 1 V (χ, φ)=− + m2χ2 + ··· (9.152) 16λ 2 2 2 with m = −2µ . We see that the eﬀect of the shift on φ1 is to produce the terms

~~1 2 2 2 µ 2 2 g v W − gvWµ∂ φ + g vW χ (9.153) 2 The last term is an interaction term and it does not play any role in the following considerations. The ﬁrst term is the mass term for the W~~

~~2 2 2 mW = g v (9.154)~~

232 whereas the second term is the mixing between the W and the Goldstone field, that we got rid of by choosing the unitary gauge. However this choice is not unique. In fact let us consider a gauge fixing of the form µ f − B =0,f= ∂ Wµ + gvξφ (9.155) where ξ is an arbitrary parameter and B(x) is an arbitrary function. By proceeding as in Section 8.3 we know that the effect of such a gauge fixing is to add to the lagrangian a term 2 1 2 1 µ 2 1 µ 2 ξ µ ξ 2 2 2 − f = − (∂ Wµ + gvξφ) = − (∂ Wµ) − gv∂ Wµφ − g v φ (9.156) 2β 2β 2β β 2β The first term is the usual covariant gauge fixing. The second term, after integration by parts reads ξ µ + gvWµ∂ φ (9.157) β and we see that with the choice β = ξ (9.158) it cancels the unwanted mixing term. Finally the third contribution 1 − ξg2v2φ2 (9.159) 2 is a mass term for the Goldstone field

2 2 2 2 mφ = ξg v = ξmW (9.160) However this mass is gauge dependent (it depends on the arbitrary parameter ξ) and this is a signal of the fact that in this gauge the Goldstone field is fictitious. In fact remember that it disappears in the unitary gauge. We will see later, in an example, that evaluating a physical amplitude the contribution of the Goldstone disappears, as it should be. Finally we have to consider the ghost contribution from the Faddeev-Popov determinant. For this purpose we need to know the variation of the gauge fixing f under an infinitesimal gauge transformation 1 δf(x)= α(x)+gvξα(x)(χ(x)+v) (9.161) g giving δf(x) 1 = δ4(x − y)+gvξ(χ(x)+v)δ4(x − y) (9.162) δα(y) g Therefore the ghost lagrangian can be written as c∗c + ξg2v2c∗c + ξg2vχc∗c (9.163) Notice that in contrast with QED, the ghosts cannot be ignored, since they are coupled to the Higgs field. We are now in the position of reading the various propagators from the quadratic part of the lagrangian

233 i • Higgs propagator (χ): k2 − m2 i • φ Goldstone propagator ( ): 2 2 k − ξmW i • c − Ghost propagator ( ): 2 2 k − ξmW The propagator for the vector ﬁeld is more involved. let us collect the various quadratic terms in Wµ

1 µν 1 2 2 1 µ 2 − Fµν F + m W − (∂ Wµ) (9.164) 4 2 W 2ξ In momentum space wave operator is 1 (gµν k2 − kµkν) − m2 gµν + kµkν (9.165) W ξ which can be rewritten as kµkν kµkν 1 gµν − (k2 − m2 )+ (k2 − ξm2 ) (9.166) k2 W k2 ξ W The propagator in momentum space is the inverse of the wave operator (except for afactor−i). To invert the operator we notice that it is written as a combination of two orthogonal projection operators P1 and P2

~~αP1 + βP2 (9.167)~~

The inverse of such an operator is simply 1 1 P + P (9.168) α 1 β 2 Therefore the W propagator is given by i kµkν iξ kµkν − gµν − − 2 − 2 2 2 − 2 2 k mW k k ξmW k µ ν − i µν − k k − = 2 2 g 2 2 (1 ξ) (9.169) k − mW k − ξmW We see that for this class of gauges the asymptotic behaviour of the propagator is 1/k2 and we can apply the usual power counting. Notice that the unitary gauge is recovered in the limit ξ →∞. In this limit the W -propagator becomes the one discussed at the beginning of the Section. Of course, the singularity of the limit destroys the asymptotic behaviour that one has for finite values of ξ.Also,for ξ →∞the ghost and the Goldstone fields decouple since their mass goes to infinity.

234 To end this Section we observe that with this approach one is able to prove at the same time that the theory is renormalizable and that it does not contain spurious states. In fact the theory is renormalizable for ﬁnite values of ξ and it does not contain spurious states for ξ →∞. But, since the gauge invariant quantities do not depend on the gauge ﬁxing, they do not depend in particular on ξ, and therefore one gets the wanted result.

~~9.6 ξ-cancellation in perturbation theory~~

In this Section we will discuss the cancellation of the ξ dependent terms in the calculation of a simple amplitude at tree level. To this end we consider again the O(2) model of the previous Section, and we add to the theory a fermion ﬁeld. We will also mimic the standard model by making transform under the gauge group only the left-handed part of the fermion ﬁeld. That is we decompose ψ as

~~ψ = ψL + ψR (9.170) where 1 − γ5 1+γ5 ψL = ψ, ψR = ψ (9.171) 2 2 and we assume the following transformation under the gauge group iα ψL → e ψL,ψR → ψR (9.172)~~

A fermion mass term would destroy the gauge invariance, since it couples left-handed andright-handedﬁelds.Infact 1 − γ5 † 1 − γ5 1+γ5 ψ¯L = ψ = ψ γ = ψ¯ (9.173) 2 2 0 2 and therefore 1+γ5 1 − γ5 ψψ¯ = ψ¯ + ψ = ψ¯LψR + ψ¯RψL (9.174) 2 2 Therefore we will introduce a Yukawa coupling between the scalar ﬁeld and the fermions, in such a way to be gauge invariant and to gives rise to a mass term for the fermion when the symmetry is spontaneously broken. We will write

∗ Lf = ψ¯LiDψˆ L + ψ¯Ri∂ψˆ R − λf (ψ¯LψRΦ+ψ¯RψLΦ ) (9.175) with Dµ = ∂µ − igWµ (9.176) Here we have used again the complex notation for the scalar ﬁeld, as in eq. (9.109), and we remind that it transforms as

~~Φ → eiαΦ (9.177)~~

~~235 Therefore Lf is gauge invariant. The lagrangian can be rewritten in terms of the ﬁeld ψ by writing explicitly the chiral projectors (1 ± γ5)/2. We get~~

1 − γ5 1+γ5 1 − γ5 µ Lf = ψi¯ ∂ˆ ψ + ψi¯ ∂ˆ ψ + gψγ¯ µ ψW 2 2 2 − λf ¯1+γ5 ¯1 γ5 − √ ψ ψ(φ1 + iφ2)+ψ ψ(φ1 − iφ2) (9.178) 2 2 2 from which − ¯ ˆ ¯ 1 γ5 µ λf ¯ ¯ L = ψi∂ψ + gψγµ ψW − √ ψψ(v + χ)+iψγ5ψφ (9.179) 2 2 where χ + v = φ1 and φ = φ2. This expression shows that the fermion ﬁeld acquires a mass through the Higgs mechanism given by

λf v mf = √ (9.180) 2 The previous expression shows also that the Higgs ﬁeld is a scalar (parity +1), whereas the Goldstone ﬁeld is a pseudoscalar (parity= -1). Consider now the fermion-fermion scattering in this model at tree level. The amplitude gets contribution from the diagrams of Fig. 9.6.1. . p' k W φ χ ++ q p k' Fig. 9.6.1 - The fermion-fermion scattering in the U(1) Higgs model.

We want to show that the ξ contribution cancels out. The ξ dependence arises from the gauge ﬁeld and the Goldstone ﬁeld exchange. Therefore we will consider only these two contributions. Let us start with the Goldstone exchange. The Feynman’s rule for the vertex is given in Fig. 9.6.2. . φ

~~λf Fig. 9.6.2 - Goldstone-fermion vertex: √ γ5 2~~

236 The corresponding amplitude is 2 λf i Mφ √ = u¯(p )γ5u(p) 2 2 u¯(k )γ5u(k) (9.181) 2 q − ξmW where q = p − p = k − k. The Feynman rule for the W-fermion vertex is given in Fig. 9.6.3. .

~~1 − γ5 Fig. 9.6.3 - W-fermion vertex: igγµ 2~~

The corresponding amplitude is − 2 1 γ5 MW =(ig) u¯(p )γµ u(p) 2 µ ν −i µν q q 1 − γ5 × − − ν 2 2 g 2 2 (1 ξ) u¯(k )γ u(k) (9.182) q − mW q − ξmW 2 It is convenient to isolate the ξ dependence by rewriting the gauge field propagator as −i qµqν 1 1 − ξ gµν − + qµqν − 2 − 2 2 2 2 − 2 q mW mW mW q ξmW −i qµqν −i qµqν gµν − == 2 2 2 + 2 2 2 (9.183) q − mW mW q − ξmW mW The first term is ξ independent and it is precisely the W propagator in the unitary gauge. The second term can be simplified by using the identity

~~µ 1 − γ5 µ 1 − γ5 q u¯(p )γµ u(p)=(p − p ) u¯(p )γµ u(p)= 2 2 1+γ5 1 − γ5 =¯u(p ) pˆ − pˆ u(p)=mf u¯(p )γ u(p) (9.184) 2 2 5 By using this relation we have~~

~~µ ν 1 − γ5 1 − γ5 q q u¯(p )γµ u(p)¯u(k )γµ u(k)= 2 2 =(mf )¯u(p )γ u(p)(−mf )¯u(k )γ u(k) 5 5 2 λf 2 = − √ v u¯(p )γ5u(p)¯u(k )γ5u(k) (9.185) 2~~

237 where we have made use of eq. (9.180). Substituting this result in MW we get − − µ ν − 2 1 γ5 i µν q q 1 γ5 MW =(ig) u¯(p )γµ u(p) g − u¯(k )γµ u(k) 2 q2 − m2 m2 2 W W 2 2 2 −i g v λf √ u p γ u p u k γ u k + 2 2 2 ¯( ) 5 ( )¯( ) 5 ( ) (9.186) q − ξmW mW 2

2 2 2 Finally, using mW = g v we see that the last term cancels exactly the Mφ amplitude. Therefore the result is the same that we would have obtained in the unitary gauge where there is no Goldstone contribution. Now let us make some comment about the result. In QED we have used the fact that the qµqν terms in the propagator do not play a physical role,since they are contracted with the fermionic current which is conserved. In the present case this does not happens, but we get a term proportional to the fermion mass. In ¯ fact the fermionic current ψγµ(1 − γ5)/2ψ it is not conserved because the fermion acquires a mass through the spontaneous breaking of the symmetry. On the other hand, the term that one gets in this way it is just the one necessary to give rise to the cancellation with the Goldstone boson contribution. It is also worth to compare what happens for diﬀerent choices of ξ.Forξ = 0 the gauge propagator is transverse −i kµkν gµν − 2 2 2 (9.187) k − mW k

This corresponds to the Landau gauge. The propagator has a pole term at k2 =0, which however it is cancelled by the Goldstone pole which also is at k2 =0.Forthe choice ξ = 1 the gauge propagator has the very simple form −igµν 2 2 (9.188) k − mW This is the Feynman-’t Hooft gauge. Here the role of the Goldstone boson, which has a propagator i 2 2 (9.189) k − mW it is to cancel the contribution of the time-like component of the gauge ﬁeld, which (0) is described by the polarization vector µ . In fact the gauge propagator can be rewritten as − µν − 3 3 − µ ν ig i (λ) (λ) i (λ) (λ) i k k gλλ 2 2 = 2 2 µ ν = 2 2 µ ν + 2 2 2 k − mW k − mW k − mW k − mW mW λ,λ=0 λ=1 (9.190) The ﬁrst piece is the propagator in the unitary gauge, whereas the second piece is cancelled by the Goldstone boson contribution.

~~238 Chapter 10~~

~~The Standard model of the electroweak interactions~~

~~10.1 The Standard Model of the electroweak interactions~~

We will describe now the Weinberg Salam model for the electroweak interactions. Let us recall that the Fermi theory of weak interactions is based on a lagrangian involving four Fermi ﬁelds. This lagrangian can be written as the product of two currents which are the sum of various pieces.

~~GF (+) µ(−) LF = √ Jµ J (10.1) 2 Limiting for the moment our considerations to the charged current for the electron and its neutrino (here ν ≡ νe)), the current is given by~~

~~† (+) ¯ (−) (+) ¯ Jµ = ψeγµ(1 − γ5)ψν ,Jµ = Jµ = ψν γµ(1 − γ5)ψe (10.2)~~

To understand the symmetry hidden in this couplings it is convenient to introduce the following spinors with 2 × 4components (ψν)L 1 − γ ψν L = = 5 (10.3) (ψe)L 2 ψe where we have introduced left-handed ﬁelds. The right-handed spinors, projected out with (1 + γ5)/2, don’t feel the weak interaction. This is, in fact, the physical meaning of the V − A interaction. By using 1+γ5 L¯ = (ψ¯ν )L, (ψ¯e)L = ψ¯ν , ψ¯e (10.4) 2

239 we can write the weak currents in the following way − (+) 1+γ5 1 γ5 0 Jµ =2ψ¯e γµ ψν =2 (ψ¯ν)L, (ψ¯e)L γµ (ψν )L 2 2 00 (ψν )L =2(ψ¯ν )L, (ψ¯e)L γµ =2Lγ¯ µτ−L (10.5) 10 (ψe)L wherewehavedeﬁned τ1 ± iτ2 τ± = (10.6) 2 the τi’s being the Pauli matrices. Analogously

(−) Jµ =2Lγ¯ µτ+L (10.7) In the ﬁfties the hypothesis of the intermediate vector bosons was advanced. Ac- cording to this idea the Fermi theory was to be regarded as the low energy limit of a theory where the currents were coupled to vector bosons, very much as QED. However, due to the short range of the weak interactions these vector bosons had to have mass. The interaction among the vector bosons and the fermions is given by g (+) −µ (−) +µ g ¯ −µ +µ Lint = √ Jµ W + Jµ W = √ Lγµ(τ−W + τ+W )L (10.8) 2 2 2 The coupling g can be related to the Fermi constant through the relation

~~2 G√F g = 2 (10.9) 2 8mW~~

Introducing real ﬁelds in place of W± W ∓ iW W ± = 1 √ 2 (10.10) 2 we get g µ µ L = Lγ¯ µ(τ W + τ W )L (10.11) int 2 1 1 2 2 Let us now try to write the electromagnetic interaction within the same formalism µ 1+γ5 1 − γ5 1 − γ5 1+γ5 µ L = eψ¯eQγµψeA = eψ¯eQ γµ + γµ ψeA em 2 2 2 2 µ = e[(ψ¯e)LQγµ(ψe)L +(ψ¯e)RQγµ(ψe)R]A (10.12) where 1+γ5 1 − γ5 (ψe)R = ψe, (ψ¯e)R = ψ¯e (10.13) 2 2 where Q = −1 is the electric charge of the electron, in unit of the electric charge of the proton, e. It is also convenient to denote the right-handed components of the electron as R =(ψe)R (10.14)

240 We can write 00 1 − τ3 (ψ¯e)Lγµ(ψe)L = Lγ¯ µ L = Lγ¯ µ L (10.15) 01 2 and (ψ¯e)Rγµ(ψe)R = Rγ¯ µR (10.16) from which, using Q = −1, 1 τ3 µ em µ L = e − Lγ¯ µL − Rγ¯ µR + Lγ¯ µ L A = ej A (10.17) em 2 2 µ From this equation we can write 1 jem = j3 + jY (10.18) µ µ 2 µ with 3 τ3 j = Lγ¯ µ L (10.19) µ 2 and Y jµ = −Lγ¯ µL − 2Rγ¯ µR (10.20) 3 These equations show that the electromagnetic current is a mixture of jµ,whichis Y a partner of the charged weak currents, and of jµ , which is another neutral current. 3 The following notations unify the charged currents and jµ

i τi j = Lγ¯ µ L (10.21) µ 2 Using the canonical anticommutators for the Fermi ﬁelds, it is not diﬃcult to prove that the charges associated to the currents i 3 i Q = d xj0(x) (10.22) span the Lie algebra of SU(2),

~~i j k [Q ,Q ]=iijkQ (10.23)~~

Y Y i The charge Q associated to the current jµ commutes with Q , as it can be easily veriﬁed noticing that the right-handed and left-handed ﬁelds commute with each other. Therefore the algebra of the charges is

~~i j k i Y [Q ,Q ]=iijkQ , [Q ,Q ] = 0 (10.24)~~

This is the Lie algebra of SU(2) ⊗ U(1), since the Qi’s generate a group SU(2), whereas QY generates a group U(1), with the two groups commuting among themselves. To build up a gauge theory from these elements we have to start with an initial lagrangian possessing an SU(2) ⊗ U(1) global invariance. This will produce

241 4 gauge vector bosons. Two of these vectors are the charged one already introduce, whereas we would like to identify one of the the other two with the photon. By evaluating the commutators of the charges with the ﬁelds we can read immediately the quantum numbers of the various particles. We have i 3 † τi τi [Lc,Q]= d x [Lc,La Lb]= Lb (10.25) 2 ab 2 cb and [R, Qi] = 0 (10.26) Y Y [Lc,Q ]=−Lc, [R, Q ]=−2R (10.27) These relations show that L transforms as an SU(2) spinor, or, as the representation 2, where we have identiﬁed the representation with its dimensionality. R belongs to the trivial representation of dimension 1. Putting everything together we have the following behaviour under the representations of SU(2) ⊗ U(1)

~~L ∈ (2, −1),R∈ (1, −2) (10.28)~~

The relation (10.18) gives the following relation among the diﬀerent charges 1 Q = Q3 + QY (10.29) em 2 As we have argued, if SU(2) ⊗ U(1) has to be the fundamental symmetry of electroweak interactions, we have to start with a lagrangian exhibiting this global symmetry. This is the free Dirac lagrangian for a massless electron and a left-handed neutrino ˆ ˆ L0 = Li¯ ∂L + Ri¯ ∂R (10.30) Mass terms, mixing left- and right-handed ﬁelds, would destroy the global symmetry. For instance, a typical mass term gives † 1+γ5 1 − γ5 1 − γ5 1+γ5 mψψ¯ = mψ γ + γ ψ = m(ψ¯RψL + ψ¯LψR) (10.31) 2 0 2 2 0 2

Weshallseelaterthatthesamemechanism giving mass to the vector bosons can be used to generate fermion masses. A lagrangian invariant under the local symmetry SU(2) ⊗ U(1) is obtained through the use of covariant derivatives µ τi i g µ µ L = Liγ¯ ∂µ − ig W + i Yµ L + Riγ¯ (∂µ + ig Y ) R (10.32) 2 µ 2

~~The interaction term is τ µ g µ µ L = gLγ¯ µ · W L − Lγ¯ µLY − g Rγ¯ µRY (10.33) int 2 2~~

242 In this expression we recognize the charged interaction with W ±. We have also, as expected, two neutral vector bosons W 3 and Y . The photon must couple to the electromagnetic current which is a linear combination of j3 and jY . The neutral vector bosons are coupled to these two currents, so we expect the photon field, 3 Aµ, to be a linear combination of Wµ and Yµ. It is convenient to introduce two orthogonal combination of these two fields, Zµ,andAµ. The mixing angle can be identified through the requirement that the current coupled to Aµ is exactly the electromagnetic current with coupling constant e. Let us consider the part of Lint involving the neutral couplings 3 µ g Y µ LNC = gj W + j Y (10.34) µ 3 2 µ By putting (θ is called the Weinberg angle)

W3 = Z cos θ + A sin θ Y = −Z sin θ + A cos θ (10.35) we get 3 1 Y µ 3 1 Y µ LNC = g sin θj + g cos θ j A + g cos θj − g sin θ j Z (10.36) µ 2 µ µ 2 µ The electromagnetic coupling is reproduced by requiring the two conditions

~~g sin θ = g cos θ = e (10.37)~~

These two relations allow us to express both the Weinberg angle and the electric charge e in terms of g and g. In the low-energy experiments performed before the LEP era (< 1990) the fundamental parameters used in the theory were e (or rather the ﬁne structure constant α)andsin2 θ. We shall see in the following how the Weinberg angle can be eliminated in favor of the mass of the Z, which is now very well known. By using eq. em 3 1 Y (10.37), and jµ = jµ + 2 jµ ,wecanwrite

~~em µ 3 em 3 µ LNC = ejµ A +[g cos θjµ − g sin θ(jµ − jµ)]Z (10.38)~~

~~Expressing g through g = g tan θ, the coeﬃcient of Zµ can be put in the form g g [g cos θ + g sin θ]j3 − g sin θjem = j3 − sin2 θjem (10.39) µ µ cos θ µ cos θ µ from which~~

~~em µ g 3 2 em µ em µ g Z µ LNC = ej A + [j − sin θj ]Z ≡ ej A + j Z (10.40) µ cos θ µ µ µ cos θ µ where we have introduced the neutral current coupled to the Z~~

~~Z 3 2 em jµ = jµ − sin θjµ (10.41)~~

~~243 The value of sin2 θ was evaluated initially from processes induced by neutral currents (as the scattering ν − e) at low energy. The approximate value is~~

~~sin2 θ ≈ 0.23 (10.42)~~

This shows that both g and g are of the same order of the electric charge. Notice that the eq. (10.37) gives the following relation among the electric charge and the couplings g and g 1 1 1 = + (10.43) e2 g2 g2 10.2 The Higgs sector in the Standard Model

According to the discussion made in the previous Section, we need three broken symmetries in order to give mass to W ± and Z.SinceSU(2) ⊗ U(1) has four generators, we will be left with one unbroken symmetry, that we should identify with the group U(1) of the electromagnetism (U(1)em), in such a way to have the corresponding gauge particle (the photon) massless. Therefore, the group U(1)em must be a symmetry of the vacuum (that is the vacuum should be electrically neutral). To realize this aim we have to introduce a set of scalar ﬁelds transforming in a convenient way under SU(2) ⊗ U(1). The simplest choice turns out to be a complex representation of SU(2) of dimension 2 (Higgs doublet). As we already noticed the vacuum should be electrically neutral, so one of the components of the doublet must also be neutral. Assume that this component is the lower one, than we will write φ+ Φ= (10.44) φ0

~~To determine the weak hypercharge, QY ,ofΦ,weusetherelation 1 Q = Q3 + QY (10.45) em 2 for the lower component of Φ. We get 1 QY (φ0)=2× 0 − − = 1 (10.46) 2~~

Since QY and Qi commute, both the components of the doublet must have the same value of QY , and using again eq. (10.45), we get 1 1 Q (φ+)= + = 1 (10.47) em 2 2 which justiﬁes the notation φ+ for the upper component of the doublet.

244 The most general lagrangian for Φ with the global symmetry SU(2) ⊗ U(1) and containing terms of dimension lower or equal to four (in order to have a renormalizable theory) is † µ 2 † † 2 LHiggs = ∂µΦ ∂ Φ − µ Φ Φ − λ(Φ Φ) (10.48) where λ>0andµ2 < 0. The potential has inﬁnitely many minima on the surface

~~µ2 v2 |Φ|2 = − = (10.49) minimum 2λ 2 with µ2 v2 = − (10.50) λ Let us choose the vacuum as the state 0 0|Φ|0 = √ (10.51) v/ 2~~

By putting + 0 φ Φ= √ + √ ≡ Φ +Φ (10.52) v/ 2 (h + iη)/ 2 0 with 0|Φ|0 = 0 (10.53) we get 1 1 |Φ|2 = v2 + vh + |φ+|2 + (h2 + η2) (10.54) 2 2 from which 1 µ4 1 1 2 V = − + λv2h2 +2λvh |φ+|2 + (h2 + η2) + λ |φ+|2 + (h2 + η2) Higgs 4 λ 2 2 (10.55) From this expression we read immediately the particle masses (φ− =(φ+)†)

2 2 2 mη = mφ+ = mφ− = 0 (10.56) and 2 2 2 mh =2λv = −2µ (10.57) It is also convenient to express the parameters appearing in the original form of the 2 2 2 2 2 potential in terms of mh e v (µ = −2mh, λ = mh/2v ). In this way we get 1 m2 m2 1 m2 1 2 V = − m2 v2 + h h2 + h h |φ+|2 + (h2 + η2) + h |φ+|2 + (h2 + η2) Higgs 8 h 2 v 2 2v2 2 (10.58) Summarizing we have three massless Goldstone bosons φ± and η, and a massive scalar h. This is called the Higgs ﬁeld.

245 As usual, by now, we promote the global symmetry to a local one by introducing the covariant derivative. Recalling that Φ ∈ (2, 1) of SU(2) ⊗ U(1), we have H g g D = ∂µ − i τ · W µ − i Yµ (10.59) µ 2 2 and H † Hµ 2 † † 2 LHiggs =(Dµ Φ) D Φ − µ Φ Φ − λ(Φ Φ) (10.60) To study the mass generation it is convenient to write Φ in a way analogous to the one used in eq. (9.127) · 0 Φ=eiξ τ/v √ (10.61) (v + h)/ 2 According to our rules the exponential should contain the broken generators. In fact there are τ1,andτ2 which are broken. Furthermore it should contain the combination of 1 and τ3 which is broken (remember that (1 + τ3)/2 is the electric charge of the doublet Φ, and it is conserved). But, at any rate, τ3 is a broken generator, so the previous expression gives a good representation of Φ around the vacuum. In fact, expanding around this state · 0 1+iξ /v i(ξ − iξ )/v 0 eiξ τ/v √ ≈ 3 1 2 √ (v + h)/ 2 i(ξ + iξ )/v 1 − iξ /v (v + h)/ 2 1 2 3 − ≈ √1 i(ξ1 iξ2) − (10.62) 2 (v + h iξ3) √ + and introducing real components for Φ (φ =(φ1 − iφ2)/ 2) 1 φ − iφ Φ=√ 1 2 (10.63) 2 v + h + iη we get the relation among the two sets of coordinates

(φ1,φ2,η,h) ≈ (ξ2, −ξ1, −ξ3,h) (10.64) By performing the gauge transformation − · Φ → e iξ τ/vΦ (10.65) τ −iξ · τ/v τ iξ · τ/v i −iξ · τ/v iξ · τ/v W µ · → e W µ · e + ∂µe e (10.66) 2 2 g

Yµ → Yµ (10.67) the Higgs lagrangian√ remains invariant in form, except for the substitution of Φ with (0, (v +h)/ 2). We will also denote old and new gauge ﬁelds with the same symbol. Deﬁning 0 1 Φd = , Φu = (10.68) 1 0

246 the mass terms for the scalar ﬁelds can be read by substituting, in the kinetic term, Φ with its expectation value v Φ → √ Φd (10.69) 2 We get g g v g − + g 3 g v −i τ · W + Y √ Φd = −i √ (τ−W + τ+W )+ τ3W + Y √ Φd 2 2 2 2 2 2 2 v g + 1 3 = −i√ √ W Φu − (gW − g Y )Φd (10.70) 2 2 2

~~Since Φd and Φu are orthogonals, the mass term is v2 g2 1 |W +|2 + (gW 3 − gY )2 (10.71) 2 2 4~~

~~From eq. (10.37) we have tan θ = g/g, and therefore~~

~~g g sin θ = , cos θ = (10.72) g2 + g2 g2 + g2 allowing us to write the mass term in the form v2 g2 g2 + g2 |W +|2 + (W 3 cos θ − Y sin θ)2 (10.73) 2 2 4~~

~~From eq. (10.35) we see that the neutral ﬁelds combination is just the Z ﬁeld, orthogonal to the photon. In this way we get~~

g2v2 1 g2 + g2 |W +|2 + v2Z2 (10.74) 4 2 4 Finally the mass of the vector bosons is given by 1 1 M 2 = g2v2,M2 = (g2 + g2)v2 (10.75) W 4 Z 4 Notice that the masses of the W ± and of the Z are not independent, since their ratio is determined by the Weinberg angle, or from the gauge coupling constants

M 2 g2 W = =cos2 θ (10.76) 2 2 2 MZ g + g Let us now discuss the parameters that we have so far in the theory. The gauge interaction introduces the two gauge couplings g and g, which can also be expressed in terms of the electric charge (or the ﬁne structure constant), and of the Weinberg angle. The Higgs sector brings in two additional parameters, the mass of the Higgs,

247 mh, and the expectation value of the ﬁeld Φ, v. The Higgs particle has not been yet discovered, and at the moment we have only an experimental lower bound on mh, from LEP, which is given by mh > 60 GeV . The parameter v can be expressed in terms of the Fermi coupling constant GF . In fact, from eq. (10.9)

~~2 G√F g = 2 (10.77) 2 8MW~~

~~2 using the expression for MW ,weget 1 v2 = √ ≈ (246 GeV )2 (10.78) 2GF~~

2 Therefore, the three parameters g, g and v can be traded for e,sinθ and GF . Another possibility is to use the mass of the Z 1 1 e2 πα M 2 = √ = √ (10.79) Z 2 2 2 4 2GF sin θ cos θ 2GF sin θ cos2 θ to eliminate sin2 θ 1 4πα sin2 θ = 1 − 1 − √ (10.80) 2 2 2GF MZ The ﬁrst alternative has been the one used before LEP. However, after LEP1, the mass of the Z is very well known

~~MZ =(91.1867 ± 0.0021) GeV (10.81)~~

Therefore, the parameters that are now used as input in the SM are α, GF and MZ . The last problem we have to solve is how to give mass to the electron, since it is impossible to construct bilinear terms in the electron field which are invariant under the gauge group. The solution is that we can build up trilinear invariant terms in the electron field and in Φ. Once the field Φ acquires a non vanishing expectation value, due to the breaking of the symmetry, the trilinear term generates the electron mass. Recalling the behaviour of the fields under SU(2) ⊗ U(1) L ∈ (2, −1),R∈ (1, −2), Φ ∈ (2, +1) (10.82) we see that the following coupling (Yukawian coupling) is invariant φ+ LY = geL¯ΦR +h.c. = ge [(ψ¯ν)L, (ψ¯e)L ] (ψe)R +h.c. (10.83) φ0 In the unitary gauge, which is obtained by using the transformation (10.65) both for Φ and for the lepton field L,weget 0 √ LY = ge [(ψ¯ν )L, (ψ¯e)L ] (ψe)R +h.c. (10.84) (v + h)/ 2

~~248 from which gev ge LY = √ (ψ¯e)L(ψe)R + √ (ψ¯e)L(ψe)Rh +h.c. (10.85) 2 2 Since~~

† 1+γ5 † † 1+γ5 1 − γ5 [(ψ¯e)L(ψe)R] =(ψ¯e ψe) = ψ γ ψe = ψ¯e ψe =(ψ¯e)R(ψe)L 2 e 2 0 2 (10.86) we get gev ge LY = √ ψ¯eψe + √ ψ¯eψeh (10.87) 2 2 Therefore the symmetry breaking generates an electron mass given by

gev me = −√ (10.88) 2 In summary, we have been able to reproduce all the phenomenological features of the V −A theory and its extension to neutral currents. This has been done with a theory that, before spontaneous symmetry breaking is renormalizable. In fact, also the Yukawian coupling has only dimension 3. The proof that the renormalizability of the theory holds also in the case of spontaneous symmetry breaking (µ2 < 0) is absolutely non trivial. In fact the proof was given only at the beginning of the seventies by ’t Hooft.

~~10.3 The electroweak interactions of quarks and the Kobayashi-Maskawa-Cabibbo matrix~~

So far we have formulated the SM for an electron-neutrino pair, but it can be extended in a trivial way to other lepton pairs. The extension to the quarks is a little bit less obvious and we will follow the necessary steps in this Section. First of all we recall that in the quark model the nucleon is made up of three quarks, − with composition, p ≈ uud, n ≈ udd. Then, the weak transition n → p + e +¯νe is obtained through − d → u + e +¯νe (10.89) and the assumption is made that quarks are point-like particles as leptons. There- fore, it is natural to write quark current in a form analogous to the leptonic current (see eqs. (10.5) and (10.7))

~~µ(±) µ Jd =2Ψ¯ Lγ τ∓ΨL (10.90) with uL 1 − γ5 u ΨL = = (10.91) dL 2 d~~

~~249 Phenomenologically, from the hyperon decays it was known that it was necessary a further current involving the quark s,givenby~~

µ(±) µ Js =2Ψ¯ Lγ τ∓ΨL (10.92) with uL 1 − γ5 u ΨL = = (10.93) sL 2 s Furthermore the total current contributing to the Fermi interaction had to be of the form µ µ µ Jh = Jd cos θC + Js sin θC (10.94) where θC is the Cabibbo angle, and

~~sin θC =0.21 ± 0.03 (10.95)~~

The presence of the angle in the current was necessary in order to explain why the weak decays strangeness violating (that is the ones corresponding to the transition + + u ↔ s,asK → µ νµ) where suppressed with respect to the decays strangeness conserving. Collecting together the two currents we get

~~µ(±) µ Jh =2Q¯Lγ τ∓QL (10.96) with uL uL QL = ≡ C (10.97) dL cos θC + sL sin θC dL and C dL = dL cos θC + sL sin θC (10.98)~~

This allows us to assign QL to the representation (2, 1/3) of SU(2) ⊗ U(1). In fact, from 1 Q = Q3 + QY (10.99) em 2 we get Y 2 1 1 Q (uL)=2( − )= (10.100) 3 2 3 and (Qem(s)=Qem(d)=−1/3) 1 1 1 QY (dC)=2(− + )= (10.101) L 3 2 3 As far as the right-handed components are concerned, we assign them to SU(2) singlets, obtaining 4 C 2 uR ∈ (1, ) d ∈ (1, − ) (10.102) 3 R 3

250 Therefore, the hadronic neutral currents contribution from quarks u, d, s are given by 3 τ3 j = Q¯Lγµ QL (10.103) µ 2 Y 1 4 2 C C j = Q¯LγµQL + u¯RγµuR − d¯ γµd (10.104) µ 3 3 3 R R 3 2 em Since the Z is coupled to the combination jµ − sin θjµ , it is easily seen that the coupling contains a bilinear term in the ﬁeld dC given by 1 1 2 µ C C 1 2 µ C C − − sin θ Z d¯ γµd + sin θZ d¯ γµd (10.105) 2 3 L L 3 R R

This gives rise to terms of the type ds¯ andsd ¯ which produce strangeness changing neutral current transitions . However the experimental result is that these transitions are strongly suppressed. In a more general way we can get this result from the simple 3 3 observation that W3 is coupled to the current jµ which has an associated charge Q which can be obtained by commuting the charges Q1 and Q2 1 2 3 3 † † 3 † [Q ,Q ]= d xd y [QLτ1QL,QLτ2QL]= d xQL[τ1,τ2]QL 3 † 3 † C† C = i d xQLτ3QL = i d x(uLuL − dL dL ) (10.106)

~~The last term is just the charge associated to a Flavour Changing Neutral Current (FCNC). We can compare the strength of a FCNC transition, which from~~

C† C † 2 † 2 † † d d = d d cos θC + s s sin θC +(d s + s d)sinθC cos θC (10.107) is proportional to sin θC cos θC , with the strength of a ﬂavour changing charged current transition as u → s, which is proportional to sin θC . From this comparison we expect the two transitions to be of the same order of magnitude. But if we 0 + − + + compare K → µ µ induced by the neutral current , with K → µ νµ induced by the charged current (see Fig. 2.3), using BR(K0 → µ+µ−), we ﬁnd

~~Γ(K0 → µ+µ−) ≤ 6 × 10−5 (10.108) Γ(K+ → µ+νµ)~~

This problem was solved in 1970 by the suggestion (Glashow Illiopoulos and Maiani, GIM) that another quark with charge +2/3 should exist, the charm. Then, one can form two left-handed doublets 1 uL 2 cL QL = C ,QL = C (10.109) dL sL where C sL = −dL sin θC + sL cos θC (10.110)

~~251 d u µ+ + µ+ K0 Z K+ W µ– ν – – µ s s~~

~~Fig. 10.3.1 - The neutral current ﬂavour changing process K0 → µ+µ− and he + + charged current ﬂavour changing process K → µ νµ.~~

~~C 3 is the combination orthogonal to dL . As a consequence the expression of Q gets modiﬁed 3 3 † † C† C C† C Q = d x (uLuL + cLcL − dL dL − sL sL ) (10.111)~~

But C† C C† C † † dL dL + sL sL = dLdL + sLsL (10.112) since the transformation matrix C dL cos θC sin θC dL C = (10.113) sL − sin θC cos θC sL is orthogonal. The same considerations apply to QY . This mechanism of cancellation turns out to be eﬀective also at the second order in the interaction, as it should 2 be, since we need a cancellation at least at the order GF in order to cope with the experimental bounds. In fact, one can see that the two second order graphs contributing to K0 → µ+µ− cancel out except for terms coming from the mass diﬀerence between the quarks u and c.Onegets

0 + − 2 2 2 A(K → µ µ ) ≈ GF (mc − mu) (10.114) allowing to get a rough estimate of the mass of the charm quark, mc ≈ 1 ÷ 3 GeV . The charm was discovered in 1974, when it was observed the bound statecc ¯ ,with J PC =1−−,knownasJ/ψ. The mass of charm was evaluated to be around 1.5 GeV . In 1977 there was the observation of a new vector resonance, the Υ, interpreted as a bound state ¯bb,whereb is a new quark, the bottom or beauty, with mb ≈ 5 GeV . Finally in 1995 the partner of the bottom, the top, t, was discovered at Fermi Lab. The mass of the top is around 175 GeV . We see that in association with three

252 leptonic doublets νe νµ ντ − , − , − (10.115) e L µ L τ L there are three quark doublets u c t , , (10.116) d L s L b L We will show later that experimental evidences for the quark b to belong to a doublet came out from PEP and PETRA much before the discovery of top. Summarizing we have three generations of quarks and leptons. Each generation includes two left-handed doublets νA uA − , A =1, 2, 3 (10.117) eA L dA L and the corresponding right-handed singlets (except for the neutrinos if we assume that they are massless). Inside each generation the total charge is equal to zero. In fact tot 2 1 Q = Qf =0− 1+3× − = 0 (10.118) i 3 3 f∈gen where we have also taken into account that each quark comes in three colors. This is very important because it guarantees the renormalizability of the SM. In fact, in general there are quantum corrections to the divergences of the currents destroying their conservation. However, the conservation of the currents coupled to the Yang- Mills fields is crucial for the renormalization properties. In the case of the SM for the electroweak interactions one can show that the condition for the absence of these corrections (Adler, Bell, Jackiw anomalies, see later) is that the total electric charge of the fields is zero. We will complete now the formulation of the SM for the quark sector, showing that the mixing of different quarks arises naturally from the fact that the quark mass eigenstates are not necessarily the same fields which couple to the gauge fields. We will denote the last ones by νA uA AL = ,qAL = ,A=1, 2, 3 (10.119) eA L dA L and the right-handed singlets by eAR,uAR,dAR (10.120) We assume that the neutrinos are massless and that they do not have any right- handed partner. The gauge interaction is then Y τ Q (fL) µ L = f¯AL i∂µ + g · W µ + g Yµ γ fAL 2 2 fL Y Q (fR) µ + f¯AR i∂µ + g γ fAR (10.121) 2 fR

253 where fAL = AL,qAL,andfAR = eAR,uAR,dAR. We recall also the weak hypercharge assignments 1 QY ( )=−1,QY (q )= AL AL 3 4 QY (e )=−2,QY (u )= AR AR 3 2 QY (d )=− (10.122) AR 3 Let us now see how to give mass to quarks. We start with up and down quarks, and a Yukawian coupling given by

~~gdq¯LΦdR +h.c. (10.123) where√ Φ is the Higgs doublet. When this takes its expectation value, Φ = (0,v/ 2), a mass term for the down quark is generated~~

gdv gdv √ d¯LdR + d¯RdL = √ dd¯ (10.124) 2 2 √ corresponding to a mass md = −gdv/ 2. In order to give mass to the up quark we need to introduce the conjugated Higgs doublet deﬁned as 0 (φ ) Φ=˜ iτ Φ = (10.125) 2 −(φ+)

~~Observing that QY (Φ)˜ = −1, another invariant Yukawian coupling is given by~~

guq¯LΦ˜uR +h.c. (10.126) √ giving mass to the up quark, mu = −guv/ 2. The most general Yukawian coupling among quarks and Higgs ﬁeld is then given by e u d LY = gAB ¯ALΦeBR + gABq¯ALΦ˜uBR + gABq¯ALΦdBR +h.c. (10.127) AB

i where gAB,withi = e, u, d, are arbitrary 3 × 3 matrices. When we shift Φ by its vacuum expectation value we get a mass term for the fermions given by v e u d ¯ Lfermion mass = √ gABe¯ALeBR + gABu¯ALuBR + gABdALdBR +h.c. (10.128) 2 AB

~~Therefore we obtain three 3 × 3 mass matrices~~

~~i v i MAB = −√ gAB,i= e, u, d (10.129) 2~~

~~254 These matrices give mass respectively to the charged leptons and to the up and down of quarks. They can be diagonalized by a biunitary transformation~~

~~i † i Md = S M T (10.130)~~

i where S and T are unitary matrices (depending on i)andMd are three diagonal matrices. Furthermore one can take the eigenvalues to be positive. This follows from the polar decomposition of an arbitrary matrix

~~M = HU (10.131) √ where H† = H = MM† is a hermitian deﬁnite positive matrix and U † = U −1 is a unitary matrix. In fact, by deﬁning~~

~~H2 = MM†,U= H−1M (10.132) we see that H is hermitian and positive deﬁnite, and that U is unitary, since~~

~~−1 U †U = M †H−2M = M †M † M −1M =1 UU† = H−1MM†H−1 = H−1H2H−1 = 1 (10.133)~~

~~Therefore, if the unitary matrix S diagonalizes H we have~~

~~† † −1 † Hd = S HS = S MU S = S MT (10.134) with T = U −1S is a unitary matrix. Each of the mass terms has the structure ψ¯LMψR.Thenweget~~

~~† † ψ¯LMψR = ψ¯LS(S MT)T ψR ≡ ψ¯LMdψR (10.135) where we have deﬁned the mass eigenstates~~

~~ψL = SψL,ψR = TψR (10.136)~~

~~We can now express the charged current in terms of the mass eigenstates. We have~~

~~+(h) u −1 d Jµ =2¯qALγµτ−qAL =2¯uALγµdAL =2¯uALγµ((S ) S )ABdBL (10.137)~~

Deﬁning V =(Su)−1Sd,V† = V −1 (10.138) we get h(+) (V ) Jµ =2¯uALγµdAL (10.139) with (V ) dAL = VABdBL (10.140) The matrix V is called the Cabibbo-Kobayashi-Maskawa (CKM) matrix, and it describes the mixing among the down type of quarks (this is conventional we could

255 have chosen to mix the up quarks as well). We see that its physical origin is that, in general, there are no relations between the mass matrix of the up quarks, M u, and the mass matrix of the down quarks, M d. The neutral currents are expressions which are diagonal in the ”primed” states, therefore their expression is the same also in the basis of the mass eigenstates, due to the unitarity of the various S matrices. In fact

~~h(3) u −1 u d −1 d jµ ≈ u¯LuL − d¯LdL =¯uL(S ) S uL − d¯L(S ) S dL =¯uLuL − d¯LdL (10.141)~~

~~For the charged leptonic current, since we have assumed massless neutrinos we get~~

~~(+) e Jµ =2¯νALγµ(S )ABeBL ≡ 2¯νALγµeAL (10.142)~~

e † where νL =(S ) νL is again a massless eigenstate. Therefore, there is no mixing in the leptonic sector, among different generations. However this is tied to our assumption of massless neutrinos. By relaxing this assumption we may generate a mixing in the leptonic sector producing a violation of the different leptonic numbers. It is interesting to discuss in a more detailed way the structure of the CKM matrix. In the case we have n generations of quarks and leptons, the matrix V , being unitary, depends on n2 parameters. However one is free to choose in an arbitrary way the phase for the 2n up and down quark fields. But an overall phase does not change V . Therefore the CKM matrix depends only on

n2 − (2n − 1) = (n − 1)2 (10.143) parameters. We can see that, in general, it is impossible to choose the phases in such awaytomakeV real. In fact a real unitary matrix is nothing but an orthogonal matrix which depends on n(n − 1) (10.144) 2 real parameters. This means that V will depend on a number of phases given by

n(n − 1) (n − 1)(n − 2) number of phases = (n − 1)2 − = (10.145) 2 2 Then, in the case of two generations V depends only on one real parameter (the Cabibbo angle). For three generations V depends on three real parameters and one phase. Since the invariance under the discrete symmetry CP implies that all the couplings in the lagrangian must be real, it follows that the SM, in the case of three generations, implies a CP violation. A violation of CP has been observed experimentally by Christensen et al. in 1964. These authors discovered that the eigenstate of CP,withCP = −1 1 |K0 = √ |K0 −|K¯ 0 (10.146) 2 2

~~256 decays into a two pion state with CP =+1withaBR~~

~~0 → + − ≈ × −3 BR(K2 π π ) 2 10 (10.147)~~

It is not clear yet, if the SM is able to explain the observed CP violation in quanti- tative terms. To complete this Section we give the experimental values for some of the matrix elements of V . The elements Vud and Vus are determined through nuclear β-, K− and hyperon-decays, using the muon decay as normalization. One gets

~~|Vud| =0.9736 ± 0.0010 (10.148) and |Vus| =0.2205 ± 0.0018 (10.149)~~

~~The elements Vcd and Vcs are determined from charm production in deep inelastic scattering νµ + N → µ + c + X. The result is~~

~~|Vcd| =0.224 ± 0.016 (10.150) and |Vcs| =1.01 ± 0.18 (10.151)~~

~~The elements Vub and Vcb are determined from the b → u and b → c semi-leptonic decays as evaluated in the spectator model, and from the B semi-leptonic exclusive decay B → D¯ ν.Onegets~~

~~|Vub| =0.08 ± 0.02 (10.152) |Vcb| and |Vcb| =0.041 ± 0.003 (10.153)~~

~~More recently from the branching ratio t → Wb, CDF (1996) has measured |Vtb|~~

~~|Vtb| =0.97 ± 0.15 ± 0.07 (10.154)~~

A more recent comprehensive analysis gives the following 90 % C.L. range for the various elements of the CKM matrix ⎛ ⎞ 0.9745 − 0.9757 0.219 − 0.224 0.002 − 0.005 V = ⎝ 0.218 − 0.224 0.9736 − 0.9750 0.036 − 0.046 ⎠ (10.155) 0.004 − 0.014 0.034 − 0.046 0.9989 − 0.9993

~~257 10.4 The parameters of the SM~~

It is now the moment to count the parameters of the SM. We start with the gauge sector, where we have the two gauge coupling constants g and g. Then, the Higgs sector is specified by the parameters µ and the self-coupling λ. We will rather use 2 2 2 v = −µ /λ and mh = −2µ . Then we have the Yukawian sector, that is that part of the interaction between fermions and Higgs field giving rise to the quark and lepton masses. If we assume three generations and that the neutrinos are all massless we have three mass parameters for the charged leptons, six mass parameters for the quarks (assuming the color symmetry) and four mixing angles. In order to test the structure of the SM the important parameters are those relative to the gauge and to the Higgs sectors. As a consequence one assumes that the mass matrix for the fermions is known. This was not the case till two years ago, before the top discovery, and its mass was unknown. The masses of the vector bosons can be expressed in terms of the previous parameters. However, the question arises about the better choice of the input parameters. Before LEP1, the better choice was to use quantities known with great precision related to the parameters (g,g,v). The choice was to use the fine structure constant 1 α = (10.156) 137.0359895(61) and the Fermi constant

−5 −2 GF =1.166389(22) × 10 GeV (10.157) as measured from the β-decay of the muon. Then, most of the experimental research of the seventies and eighties was centered about the determination of sin2 θ.The other parameters as (mt,mh) aﬀect only radiative corrections, which, in this type of experiments can be safely neglected due to the relatively large experimental errors. 2 Therefore, in order to relate the set of parameters (α, GF , sin θ)to(g,g ,v)wecan use the tree level relations. The situation has changed a lot after the beginning of running of LEP1. In fact, the mass of the Z has been measured with great precision, −5 δMZ /MZ ≈ 2 × 10 . In this case the most convenient set is (α, GF ,MZ ), with

~~MZ =(91.1867 ± 0.0021) GeV (10.158)~~

Also the great accuracy of the experimental measurements requires to take into account the radiative corrections. We will discuss this point in more detail at the end of these lectures. However we notice that one gets informations about parameters as (mt,mh) just because they aﬀect the radiative corrections. In particular, the radiative corrections are particularly sensitive to the top mass. In fact, LEP1 was able to determine the top mass, in this indirect way, before the top discovery.

~~258 10.5 Ward identities and anomalies~~

Up to now we have assumed that the symmetries for the classical lagrangian are still valid at the quantum level. This is not always so and the motivations can be easily understood. The problem has to do with the invariance of the functional integration measure with respect to the symmetry transformation. In fact, we will see that there are situations in which the measure is not invariant and therefore the symmetry is broken at the quantum level. Before dealing with this problem it will be convenient to rephrase the Noether’s theorem in a way more suitable to derive the Ward identities in a quantum ﬁeld theory. These identities are nothing but the expression of the symmetry in the quantum context. Let us consider a ﬁeld theory with a lagrangian invariant under the following global transformation

φi → φi + δφi (10.159) with A δφi(x)=−iA(T )ijφj(x) (10.160) Then, let us consider the transformation of the action under the same transformation of the fields but with A → A(x) a space-time function. We find L L 4 ∂ ∂ δS = d x δφi + δφi,µ (10.161) ∂φi ∂φi,µ where A A δφi,µ = −iA(T )ij∂µφj − i(∂µA)(T )ijφj (10.162) Substituting we find L L 4 ∂ A ∂ A δS = d x − i A(T )ijφj − i A(T )ij∂µφj ∂φi ∂φi,µ ∂L A − i (T )ijφj∂µA (10.163) ∂φi,µ

Since for A independent on x we have δS = 0 by hypothesis, it follows that the ﬁrst two terms must vanish. Therefore we are left with 4 µ A δS = d x∂ Ajµ (10.164) where A ∂L A jµ = −i (T )ijφj (10.165) ∂φi,µ are the Noether’s currents after factorizing the inﬁnitesimal parameters A. However, if we take A(x) to describe a variation around the classical solutions, that is if the

259 ﬁelds inside the currents satisfy the equations of motion, then we must have δS =0. Then, writing 4 µ δS = − d xA∂µjA (10.166) we get µ ∂µjA = 0 (10.167) Let us now consider the generating functional associated to this theory 4 iS[φ]+i d xηiφi Z[η]= D(φ)e (10.168)

~~If we perform the change of variables~~

A φi → φi + δφi = φi − iA(x)(T )ijφj (10.169) and assume that the integration measure is invariant, we get 4 4 µ A A iS[φ]+i d xηiφi i d x( ∂ Ajµ + Aηi(T )ijφj) Z[η]= D(φ)e e (10.170)

By expanding this expression at the ﬁrst order in A(x) and using the fact that the previous expression must agree with eq. (10.168) for any A(x), we get 4 iS[φ]+i d xηiφi µ A A 0= D(φ)e −i∂ jµ + ηi(T )ijφj (10.171)

~~From this equation we can generate the Ward identities by diﬀerentiating with respect to ηi and then putting ηi = 0. For instance, for ηi =0weget~~

~~µ A ∂ 0|jµ (x)|0 = 0 (10.172)~~

~~By diﬀerentiating once (remember that one gets the T ∗ product) we have~~

~~This derivation is correct only if the integration measure is invariant under the change of variables (10.169). We will illustrate a simple situation where the measure~~

260 is not invariant. Consider a massless fermion in a gauge field. For the following considerations the gauge field can be taken as an external field, but the same applies also to the quantized case. In fact for an external gauge field we define the generating functional as i d4x ψi¯ Dψˆ Z = DψDψe¯ (10.175) where Dµ = ∂µ + igAµ (10.176) As we will be interested only in checking the current conservation we can put to zero the external sources for the fermion fields (see the previous derivation). If the gauge field Aµ is quantized we have to insert a further functional integration in Aµ but this will not play any role in the following, so for simplicity we will take Aµ as an external field. The lagrangian is invariant under the global transformation

~~ψ → eiαγ5ψ (10.177) giving rise to the classically conserved current ¯ Jµ = ψγµγ5ψ (10.178)~~

~~We will show that the quantum corrections destroy the conservation law. To this end we proceed as before by performing the change of variables inside Z~~

~~ψ(x) → ψ (x)=(1+iα(x)γ5)ψ(x) ¯ ¯ ¯ ψ(x) → ψ (x)=ψ(x)(1 + iα(x)γ5) (10.179)~~

~~The variation of the action is given by 4 ¯ µ δS = d xα(x)∂µ(ψγ γ5ψ) (10.180)~~

~~In order to evaluate the change of the integration measure it is convenient to expand the fermion ﬁeld in a basis of eigenvectors of Dˆ~~

~~iDφˆ m(x)=λmφm(x) µ −iDµφ˜m(x)γ = λmφ˜m(x) (10.181) where we have used a discrete notation for the eigenvalues. In the case of a zero gauge ﬁeld Aµ the eigenvalues λm are given by~~

~~2 2 −||2 λm = k0 k (10.182) Notice that they become negative deﬁnite after being Wick rotated~~

~~2 →− 2 −||2 − 2 λm k4 k = kE (10.183)~~

261 By hypothesis the eigenvectors φm span an orthonormal basis and therefore we can expand ψ and ψ¯ as ψ(x)= amφm(x), ψ¯(x)= bmφ˜m(x) (10.184) m m where am and bm are Grassmann coefficients. Then, except for a possible constant we have DψDψ¯ = damdbm (10.185) m We can evaluate the effect of the change of variables on the Grassmann coefficients by starting from ψ (x)=(1+iα(x)γ5)ψ(x)= amφm(x) (10.186) m

Then, from the orthogonality relation 4 † d xφmφn = δmn (10.187) we get 4 † 4 † am = d xφm(x)ψ (x)=am + d xφm(x)iα(x)γ5φn(x)an = am + Cmnan n n (10.188) with 4 † Cmn = d xφm(x)iα(x)γ5φn(x) (10.189)

A similar transformation holds for bm. Therefore, from the transformation rules for the Grassmann measure we get 1 DψDψ¯ = DψDψ¯ (10.190) det|I|2 where I =1+C is the jacobian of the transformation. Since we are interested in the inﬁnitesimal transformation, we have

~~det|I| = eTr log(1 + C) ≈ eTrC (10.191) that is 4 † log det|I| ≈ TrC = i d xα(x) φn(x)γ5φn(x) (10.192) n~~

The coefficient of α(x) is the trace of γ5 taken over all the Hilbert space. Although the trace of γ5 is zero on the spinor space it gets multiplied by an infinite factor coming from the total Hilbert space. Therefore this expression is ill defined and it

262 needs to be regularized. Furthermore, we would like to maintain the gauge invariance, and therefore we should regulate the expression in a gauge invariant way. A natural choice for the regularization is the following

~~2 2 † † λn/M φn(x)γ5φn(x) = lim φn(x)γ5φn(x)e (10.193) M→∞ n n~~

This is in fact a convergence factor in the euclidean space. The previous expression can be rewritten as † † (iDˆ)2/M 2 φn(x)γ5φn(x) = lim φn(x)γ5e φn(x) M→∞ n n (iDˆ)2/M 2 = lim x|tr γ5e |x (10.194) M→∞ where tr is the trace over the Dirac matrices. The evaluation of Dˆ)2 gives

2 µ ν 1 µ ν 1 µ ν (Dˆ) = γµγνD D = [γµ,γν] D D + [γµ,γν]−D D 2 + 2 2 i µ ν 2 g µν = D − σµν [D ,D ]− = D + σµν F (10.195) 2 2 In order to get a contribution from the trace over the Dirac indices we need at least four γ matrices. Therefore the leading term is obtained by expanding the µν 2 exponential up to the order (σµν F ) and neglecting Aµ in all the other places. Therefore we get 2 † 1 g µν −/M 2 φn(x)γ5φn(x) = lim tr γ5 − σµν F x|e |x (10.196) M→∞ 2 n 2! 2M

The matrix element can be easily evaluated by inserting momentum eigenstates and performing a Wick’s rotation 4 4 4 − 2 d k 2 2 − − d kE − 2 2 M x|e /M |x = lim ek /M e ik(x y) = i e kE/M = i x→y (2π)4 (2π)4 16π2 (10.197) Therefore 2 † ig µν ρλ φ γ φn = tr[γ σµν σρλ]F F (10.198) n 5 · 2 5 n 8 16π In our deﬁnitions 0 1 2 3 γ5 = iγ γ γ γ (10.199) The trace is easily evaluated by calculating ﬁrst

~~0 1 2 3 tr[γ5σ01σ23]=−itr[γ γ γ γ γ0γ1γ2γ3]=−4i (10.200)~~

263 and using 0123 =1weget tr[γ5σµν σρλ]=4iµνρλ (10.201) Then 2 † g µν ρλ n − µνρλ φnγ5φ = 2 F F (10.202) n 32π Therefore the determinant we are looking for is given by 2 4 g µν ρλ −i d xα(x) µνρλF F det |I| = e 32π2 (10.203) and after the change of variables the generating functional is given by 2 4 ¯ 4 ¯ µ g µν ρλ i d x ψiDψˆ i d xα(x)(∂µ(ψγ γ5ψ)+ µνρλF F ) Z = DψDψe¯ e 16π2 (10.204) leading to the anomalous conservation law

2 µ g µν ρλ ∂µj = − µνρλF F (10.205) 5 16π2 This is called the Adler-Bell-Jackiw (ABJ) anomaly. The previous result can be extended to the case of non-abelian chiral symmetries as, for instance, in the standard model where the SU(2) symmetry acts only on the left-handed fermions and therefore the corresponding transformations involve γ5. The ABJ anomaly is modiﬁed by a factor

~~A B C Tr[T [T ,T ]+] (10.206)~~

A where the term T comes in the chiral transformation in conjunction with γ5, whereas the other two matrices arise from the regulator. It is a crucial point in the renormalization of the gauge theories that the gauge symmetry is preserved at the quantum level. It follows that the standard model is consistent only if the anomaly vanishes. One has to check that the factor in eq. (10.206) vanishes for all the generators of SU(2) ⊗ U(1). It is not diﬃcult to show that this is indeed the case only if the sum of the electric charges of the fermions is zero Q = 0 (10.207) fermions This condition holds for any generation, in fact 2 1 Q + Q +3(Q + Q )=−1+0+3 − = 0 (10.208) electron neutrino up down 3 3

~~264 Appendix A~~

~~A.1 Properties of the real antisymmetric matrices~~

Given a real and antisymmetric matrix n × nA, we want to show that it is possible to put it in the form (7.61) by means of an orthogonal transformation. To this end, let us notice that iA is a hermitian matrix and therefore it can be diagonalized through a unitary transformation U:

~~† U(iA)U = Ad (A.1) where Ad is diagonal and real. The eigenvalues of iA satisfy the equation~~

~~det|iA − λ · 1| =0 (A.2)~~

~~Since AT = −A it follows~~

~~det|iA − λ · 1| =det|(iA − λ · 1)T | =det|−iA − λ · 1| =0 (A.3)~~

Therefore, if λ is an eigenvalue, the same is for −λ. lo e’. It follows that Ad can be written in the form ⎡ ⎤ λ1 000··· ⎢ ⎥ ⎢ 0 −λ1 00···⎥ ⎢ ⎥ ⎢ 00λ2 0 ···⎥ ⎢ ⎥ Ad = ⎢ 000−λ2 ···⎥ (A.4) ⎢ ⎥ ⎢ ·······⎥ ⎣ ·······⎦ ·······

Notice that if n is odd, Ad has necessarily a zero eigenvalue, that we will write as (Ad)nn. We will assume also to order the matrix in such a way that all the λi are positive.Each submatrix of the type λi 0 (A.5) 0 −λi

265 can be put in the form 0 iλi (A.6) −iλi 0 through the unitary 2 × 2 matrix 1 i 1 V2 = √ (A.7) 2 1 i In fact λi 0 † 0 iλi V2 V2 = (A.8) 0 −λi −iλi 0 Deﬁning ⎡ ⎤ V2 0 ··· ⎢ ⎥ ⎢ 0 V2 ···⎥ ⎢ ⎥ V = ⎢ · · ···⎥ (A.9) ⎣ · · ···⎦ · · ··· with Vnn =0forn odd, we get ⎡ ⎤ 0 iλ1 00··· ⎢ ⎥ ⎢ −iλ1 00 0···⎥ ⎢ ⎥ ⎢ 000iλ20 ···⎥ † ⎢ ⎥ VAdV = ⎢ 00−iλ1 0 ···⎥ = iAs (A.10) ⎢ ⎥ ⎢ ··· ····⎥ ⎣ ··· ····⎦ ··· ···· with As deﬁned in (7.61). Therefore

~~† (VU)(iA)(VU) = iAs (A.11) or † (VU)(A)(VU) = As (A.12)~~

~~But A and As are real matrices, and therefore also VU must be real. However being VU real and unitary it must be orthogonal.~~

~~266~~